Derivatives and Second-Order Derivatives for CUET Maths

Q: How do you differentiate an implicit function?

Differentiate every term of the equation with respect to x, treating y as a function of x, so that each term with y picks up a factor dy/dx by the chain rule. Then collect the dy/dx terms and solve for it.

Q: How do you find dy/dx for a parametric function?

Differentiate x and y separately with respect to the parameter, then divide: dy/dx = (dy/dt)/(dx/dt), wherever dx/dt ≠ 0. The answer comes out in terms of t.

Q: What is a second-order derivative?

It is the derivative of the derivative: differentiate y to get dy/dx, then differentiate that again. It is written d^2y/dx^2, y'' or f''(x), and it tells how fast the rate of change is itself changing.

Q: How do you find the second derivative of a parametric function?

Differentiate dy/dx, which is in terms of t, with respect to t, and divide by dx/dt once more: d^2y/dx^2 = d/dt(dy/dx) ÷ dx/dt. It is not d^2y/dt^2 divided by d^2x/dt^2.

Q: How do you show that y satisfies an equation with its second derivative?

Find dy/dx and d^2y/dx^2, put all three into the expression, and simplify; the answer is usually 0 or a simple multiple of y. Two patterns cover most questions:

About 5 min to learn · 96 questions

Derivatives are part of Section A of CUET Maths, the section every candidate answers, whether you take Mathematics or Applied Mathematics, and they are the tool behind the next three chapters: increasing functions, maxima and minima, and integration. Learn each rule here, from the chain rule to implicit and parametric functions and second-order derivatives, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section A1

On this page · 10 sections

01The rules of differentiation
02Implicit functions
03Parametric functions
04Second-order derivatives
05Equations with second derivatives
06How CUET asks this chapter
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

The rules of differentiation

A ball thrown straight up is at height $h = 20t - 5t^2$ metres after $t$ seconds. Its derivative, $\frac{dh}{dt} = 20 - 10t$ , is the ball's velocity: 10 m/s upwards after 1 second, and 0 at the top, after 2 seconds. This chapter is about finding such derivatives quickly, and then the derivative of the derivative, which for the ball turns out to be gravity.

Every question starts from a short list of standard derivatives.

Function	Derivative	Function	Derivative
$x^n$	$nx^{n-1}$	$e^x$	$e^x$
$\sin x$	$\cos x$	$a^x$	$a^x\log a$
$\cos x$	$-\sin x$	$\log x$	$\frac{1}{x}$
$\tan x$	$\sec^2 x$	$\cot x$	$-\operatorname{cosec}^2 x$
$\sec x$	$\sec x\tan x$	$\operatorname{cosec} x$	$-\operatorname{cosec} x\cot x$

The derivatives of the "co" functions, $\cos$ , $\cot$ and $\operatorname{cosec}$ , all carry a minus sign. Here $\log x$ is the natural logarithm, to base $e$ .

Product and quotient rules

(uv)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

What is the chain rule?

The chain rule differentiates a function of a function: take the derivative of the outer function, leaving the inside as it is, then multiply by the derivative of the inside. In symbols, if $y = f(u)$ and $u = g(x)$ , then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Example 1 · The chain rule

Differentiate $y = (2x^2 + 1)^4$ .

The outer function is the fourth power: its derivative is $4(2x^2 + 1)^3$ .
The inside, $2x^2 + 1$ , has derivative $4x$ .

Answer

\frac{dy}{dx} = 4(2x^2 + 1)^3 \cdot 4x = 16x(2x^2 + 1)^3

Watch out

Most wrong options are the outer derivative alone, with the inside's derivative left out. With several layers, such as $\sin^2 3x$ , one factor comes from each layer: the square, the sine and the $3x$ .

Implicit functions

Some curves are not written as $y = f(x)$ . On the circle $x^2 + y^2 = 36$ , $y$ still depends on $x$ , but it is given implicitly.

How do you differentiate an implicit function?

Differentiate every term of the equation with respect to $x$ , treating $y$ as a function of $x$ , so that each term with $y$ picks up a factor $\frac{dy}{dx}$ by the chain rule. Then collect the $\frac{dy}{dx}$ terms and solve for it.

So $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ , $\frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx}$ , and a term like $xy$ needs the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ .

Example 2 · An implicit derivative

Find $\frac{dy}{dx}$ for $x^2y + y^3 = 10$ , and its value at $(3, 1)$ .

Differentiate: $2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$ , using the product rule on $x^2y$ .
Collect: $(x^2 + 3y^2)\frac{dy}{dx} = -2xy$ , so $\frac{dy}{dx} = -\frac{2xy}{x^2 + 3y^2}$ .
At $(3, 1)$ : $-\frac{2(3)(1)}{9 + 3}$ .

Answer

$\frac{dy}{dx} = -\frac{1}{2}$ at $(3, 1)$

Watch out

Check what is asked. Solving for the reciprocal gives $\frac{dx}{dy}$ , and it is often one of the options.

Parametric functions

A curve can also be described by a third variable, a parameter $t$ : $x = f(t)$ and $y = g(t)$ . As $t$ changes, the point $(x, y)$ moves along the curve.

How do you find dy/dx for a parametric function?

Differentiate $x$ and $y$ separately with respect to the parameter, then divide: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , wherever $\frac{dx}{dt} \ne 0$ . The answer comes out in terms of $t$ .

Example 3 · An ellipse

For $x = 5\cos t$ , $y = 3\sin t$ , find $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$ .

$\frac{dx}{dt} = -5\sin t$ and $\frac{dy}{dt} = 3\cos t$ .
$\frac{dy}{dx} = \frac{3\cos t}{-5\sin t} = -\frac{3}{5}\cot t$ .

Answer

At $t = \frac{\pi}{4}$ , $\cot t = 1$ , so $\frac{dy}{dx} = -\frac{3}{5}$ .

Second-order derivatives

What is a second-order derivative?

It is the derivative of the derivative: differentiate $y$ to get $\frac{dy}{dx}$ , then differentiate that again. It is written $\frac{d^2y}{dx^2}$ , $y''$ or $f''(x)$ , and it tells how fast the rate of change is itself changing.

For the ball, $\frac{dh}{dt} = 20 - 10t$ , so $\frac{d^2h}{dt^2} = -10$ : the velocity drops by 10 m/s every second, all the way up and all the way down. That is the pull of gravity.

Example 4 · A second derivative

Find $\frac{d^2y}{dx^2}$ for $y = xe^{2x}$ .

Product rule: $\frac{dy}{dx} = e^{2x} + 2xe^{2x} = (1 + 2x)e^{2x}$ .
Again: $\frac{d^2y}{dx^2} = 2e^{2x} + (1 + 2x) \cdot 2e^{2x}$ .

Answer

\frac{d^2y}{dx^2} = 4(1 + x)e^{2x}

Watch out

$\frac{d^2y}{dx^2}$ is not $\left(\frac{dy}{dx}\right)^2$ . For $y = x^3$ , the second derivative is $6x$ , while the square of the first is $9x^4$ .

How do you find the second derivative of a parametric function?

Differentiate $\frac{dy}{dx}$ , which is in terms of $t$ , with respect to $t$ , and divide by $\frac{dx}{dt}$ once more: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}$ . It is not $\frac{d^2y}{dt^2}$ divided by $\frac{d^2x}{dt^2}$ .

Example 5 · A parametric second derivative

For $x = 2t$ , $y = t^3$ , find $\frac{d^2y}{dx^2}$ at $t = 2$ .

$\frac{dy}{dx} = \frac{3t^2}{2}$ .
$\frac{d}{dt}\left(\frac{3t^2}{2}\right) = 3t$ , and dividing by $\frac{dx}{dt} = 2$ gives $\frac{d^2y}{dx^2} = \frac{3t}{2}$ .

Answer

$\frac{d^2y}{dx^2} = 3$ at $t = 2$ . The wrong rule would divide $6t$ by $\frac{d^2x}{dt^2} = 0$ .

For an implicit curve, differentiate the expression for $\frac{dy}{dx}$ again, then replace $\frac{dy}{dx}$ by what you found. On the circle $x^2 + y^2 = 36$ , $\frac{dy}{dx} = -\frac{x}{y}$ and $\frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3} = -\frac{36}{y^3}$ .

Equations with second derivatives

A favourite question gives $y$ with some constants and asks which equation $y$ , $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ satisfy together.

How do you show that y satisfies an equation with its second derivative?

Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ , put all three into the expression, and simplify; the answer is usually 0 or a simple multiple of $y$ . Two patterns cover most questions:

$y = A\sin kx + B\cos kx$ gives $\frac{d^2y}{dx^2} = -k^2y$ , so $\frac{d^2y}{dx^2} + k^2y = 0$ ;
$y = Ae^{kx} + Be^{-kx}$ gives $\frac{d^2y}{dx^2} = k^2y$ .

Example 6 · Two exponentials

Show that $y = 3e^{2x} + 2e^{-x}$ satisfies $\frac{d^2y}{dx^2} - \frac{dy}{dx} - 2y = 0$ .

$\frac{dy}{dx} = 6e^{2x} - 2e^{-x}$ and $\frac{d^2y}{dx^2} = 12e^{2x} + 2e^{-x}$ .
$\left(12e^{2x} + 2e^{-x}\right) - \left(6e^{2x} - 2e^{-x}\right) - 2\left(3e^{2x} + 2e^{-x}\right)$ collects to $(12 - 6 - 6)e^{2x} + (2 + 2 - 4)e^{-x}$ .

Answer

Both brackets are 0, so the expression is 0.

Key idea

For $y = Ae^{mx} + Be^{nx}$ , the equation is always $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny = 0$ , because $m$ and $n$ are the roots of $r^2 - (m + n)r + mn = 0$ . In the example $m = 2$ and $n = -1$ , giving $-1$ and $-2$ .

How CUET asks this chapter

Rules: the derivative of a product, quotient or composite function, sometimes from given values such as $g(2)$ and $g'(2)$ .
Implicit and parametric: $\frac{dy}{dx}$ as an expression or at a point, with $\frac{dx}{dy}$ among the options.
Second derivatives: $\frac{d^2y}{dx^2}$ or $f''(a)$ for a given function, including a parametric one.
Relations: the value of $k$ in $\frac{d^2y}{dx^2} + ky = 0$ , or which equation a given $y$ satisfies, often as a statement or match-the-list question.

Exam shortcuts

$\frac{d}{dx}f(ax + b) = af'(ax + b)$ : the inner factor is just $a$ .
$\frac{d^2}{dx^2}e^{kx} = k^2e^{kx}$ ; $\frac{d^2}{dx^2}\sin kx = -k^2\sin kx$ , and the same for $\cos kx$ .
Parametric: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ; second derivative: divide by $\frac{dx}{dt}$ once more.
Circle $x^2 + y^2 = r^2$ : $\frac{dy}{dx} = -\frac{x}{y}$ and $\frac{d^2y}{dx^2} = -\frac{r^2}{y^3}$ .
$y = Ae^{mx} + Be^{nx}$ : $y'' - (m + n)y' + mny = 0$ .

Common mistakes

Dropping the derivative of the inside in the chain rule.
Writing $(uv)' = u'v'$ , or the quotient rule's numerator the wrong way round.
Forgetting $\frac{dy}{dx}$ on a $y$ term when differentiating implicitly.
Dividing the wrong way and giving $\frac{dx}{dy}$ .
Squaring the first derivative instead of differentiating it again.
For a parametric second derivative, stopping before the second division by $\frac{dx}{dt}$ .

Quick revision

Rules

$(uv)' = u'v + uv'$ ; $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
Chain rule: outer derivative times inner derivative
$\cos$ , $\cot$ , $\operatorname{cosec}$ : minus signs

Implicit and parametric

Implicit: each $y$ term picks up $\frac{dy}{dx}$
Parametric: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Second order

$\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ , not its square
Parametric: $\frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}$
$\sin kx$ , $\cos kx$ : $y'' = -k^2y$ ; $e^{\pm kx}$ : $y'' = k^2y$

The sign of the first derivative tells where a function rises and falls: increasing and decreasing functions.

Next · 09Quick revision

Check your understanding

96 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

The rules of differentiationQ1 of 96Q1

The rules of differentiation

28 questions

Question 1

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What is $\frac{d}{dx}\left(\left(3 x + 2\right)^{5}\right)$ ?

5 \left(3 x + 2\right)^{4}

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15 \left(3 x + 2\right)^{4}

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- 15 \left(3 x + 2\right)^{4}

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30 \left(3 x + 2\right)^{4}

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AnswerB · $15 \left(3 x + 2\right)^{4}$

Use the chain rule: differentiate the outer function, then multiply by the derivative of what is inside it.
$\frac{d}{dx}\left(\left(3 x + 2\right)^{5}\right) = 15 \left(3 x + 2\right)^{4}$ .

Worth noting

$5 \left(3 x + 2\right)^{4}$ differentiates the outside only and drops the inside's derivative.

Question 2

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Differentiate $x^{2} \sin x$ with respect to $x$ .

2 x \cos x

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x^{2} \cos x

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x^{2} \cos x + 2 x \sin x

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- x^{2} \cos x + 2 x \sin x

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AnswerC · $x^{2} \cos x + 2 x \sin x$

Product rule: $(uv)' = u'v + uv'$ , with $u = x^{2}$ and $v = \sin x$ .
$(2 x)(\sin x) + (x^{2})(\cos x) = x^{2} \cos x + 2 x \sin x$ .

Worth noting

$2 x \cos x$ multiplies the two derivatives, which is not the product rule.

Question 3

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Differentiate $\sec x$ .

\tan x \sec x

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- \tan x \sec x

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\tan x

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\left(2 \tan^{2} x + 1\right) \sec x

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AnswerA · $\tan x \sec x$

$\frac{d}{dx}\left(\sec x\right) = \tan x \sec x$ , a standard derivative.
Keep the sign: the derivatives of the co-functions (cos, cot, cosec) are negative.

Worth noting

$- \tan x \sec x$ has the wrong sign.

Question 4

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Differentiate $\log\left(x^{2} + 4\right)$ with respect to $x$ .

\frac{1}{x^{2} + 4}

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\frac{2 x}{x^{2} + 4}

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\frac{1}{2 x}

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2 x \log{\left(x^{2} + 4 \right)}

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AnswerB · $\frac{2 x}{x^{2} + 4}$

$\frac{d}{dx}\log u = \frac{1}{u}\frac{du}{dx}$ , with $u = x^{2} + 4$ .
$\frac{1}{x^{2} + 4} \cdot 2x = \frac{2x}{x^{2} + 4}$ .

Worth noting

$\frac{1}{x^{2} + 4}$ stops at the logarithm and leaves out the derivative of $x^{2} + 4$ .

Question 5

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What is $\frac{d}{dx}\left(e^{\sin x}\right)$ ?

e^{\sin x}

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e^{\cos x}

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e^{\sin x - 1} \sin x

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e^{\sin x} \cos x

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AnswerD · $e^{\sin x} \cos x$

$\frac{d}{dx}e^{u} = e^{u}\frac{du}{dx}$ , with $u = \sin x$ .
$e^{\sin x} \cdot \cos x$ .

Worth noting

$e^{\sin x}$ alone treats the power as if it were $x$ ; the derivative of $\sin x$ must multiply it.

Question 6

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Differentiate $\log(\sin x)$ with respect to $x$ .

\cot x

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\frac{1}{\sin x}

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\tan x

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- \cot x

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AnswerA · $\cot x$

By the chain rule, $\frac{d}{dx}\log u = \frac{1}{u}\frac{du}{dx}$ with $u = \sin x$ .
$\frac{1}{\sin x} \cdot \cos x = \cot x$ .

Worth noting

$\frac{1}{\sin x}$ stops at the logarithm and forgets to multiply by the derivative of $\sin x$ .

Question 7

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$g(2) = 3$ , $g'(2) = -1$ , $h(2) = 4$ and $h'(2) = 5$ . If $f(x) = g(x)h(x)$ , what is $f'(2)$ ?

-5

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10

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11

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12

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AnswerC · $11$

$f'(2) = g'(2)h(2) + g(2)h'(2)$ .
$= (-1)(4) + (3)(5) = -4 + 15 = 11$ .

Worth noting

$-5$ is $g'(2)h'(2)$ , the product of the two derivatives.

Question 8

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What is the derivative of $\frac{x + 1}{x - 1}$ ?

- \frac{2}{\left(x - 1\right)^{2}}

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\frac{2}{\left(x - 1\right)^{2}}

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1

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\frac{2 x}{\left(x - 1\right)^{2}}

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AnswerA · $- \frac{2}{\left(x - 1\right)^{2}}$

Quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ , with $u = x + 1$ and $v = x - 1$ .
This simplifies to $- \frac{2}{\left(x - 1\right)^{2}}$ .

Worth noting

$\frac{2}{\left(x - 1\right)^{2}}$ has the numerator the wrong way round: it is $u'v - uv'$ , not $uv' - u'v$ .

Question 9

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What is $\frac{d}{dx}\left(5^x\right)$ ?

x \cdot 5^{x - 1}

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5^{x} \log 5

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5^{x}

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\frac{5^{x}}{\log 5}

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AnswerB · $5^{x} \log 5$

Write $5^x = e^{x \log 5}$ .
By the chain rule its derivative is $e^{x \log 5} \cdot \log 5 = 5^x \log 5$ .

Worth noting

$x \cdot 5^{x - 1}$ treats $5^x$ like a power of $x$ ; here the variable is in the exponent.

Question 10

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If $f(x) = \left(x^{2} + 1\right)^{3}$ , what is $f'(1)$ ?

8

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12

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24

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48

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AnswerC · $24$

By the chain rule, $f'(x) = 3(x^2 + 1)^2 \cdot 2x = 6 x \left(x^{2} + 1\right)^{2}$ .
At $x = 1$ : $6(1)(2)^2 = 24$ .

Worth noting

8 is $f(1)$ , the value of the function itself; 12 forgets the inner factor $2x$ .

Question 11

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$g(1) = 3$ , $g'(1) = 2$ and $f'(3) = 4$ . What is the derivative of $f(g(x))$ at $x = 1$ ?

4

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6

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7

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8

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AnswerD · $8$

By the chain rule, $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$ .
At $x = 1$ : $f'(g(1)) \cdot g'(1) = f'(3) \cdot 2 = 4 \times 2 = 8$ .

Worth noting

4 is $f'(3)$ alone; the inner function's derivative $g'(1) = 2$ still multiplies it.

Question 12

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Which of the following statements are true?

I. $\frac{d}{dx}(uv) = \frac{du}{dx} \cdot \frac{dv}{dx}$ .

II. $\frac{d}{dx}\left(e^{2x}\right) = 2e^{2x}$ .

III. $\frac{d}{dx}(\log x) = \frac{1}{x}$ for $x > 0$ .

IV. $\frac{d}{dx}(\cos x) = \sin x$ .

II and III onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answerIII onlyCorrect answerYour answer

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AnswerA · II and III only

II is the chain rule on $e^{2x}$ , and III is a standard derivative, so both are true.
I is false: $(uv)' = u'v + uv'$ . IV is false: $\frac{d}{dx}(\cos x) = -\sin x$ .

Question 13

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What is $\frac{d}{dx}\left(\cos{\left(3 x^{2} \right)}\right)$ ?

- \sin{\left(3 x^{2} \right)}

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6 x \sin{\left(3 x^{2} \right)}

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- 12 x \sin{\left(3 x^{2} \right)}

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- 6 x \sin{\left(3 x^{2} \right)}

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AnswerD · $- 6 x \sin{\left(3 x^{2} \right)}$

Use the chain rule: differentiate the outer function, then multiply by the derivative of what is inside it.
$\frac{d}{dx}\left(\cos{\left(3 x^{2} \right)}\right) = - 6 x \sin{\left(3 x^{2} \right)}$ .

Worth noting

$- \sin{\left(3 x^{2} \right)}$ differentiates the outside only and drops the inside's derivative.

Question 14

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Differentiate $e^{x} \cos x$ with respect to $x$ .

- e^{x} \sin x

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- e^{x} \sin x + e^{x} \cos x

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e^{x} \sin x + e^{x} \cos x

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e^{x} \cos x

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AnswerB · $- e^{x} \sin x + e^{x} \cos x$

Product rule: $(uv)' = u'v + uv'$ , with $u = e^{x}$ and $v = \cos x$ .
$(e^{x})(\cos x) + (e^{x})(- \sin x) = - e^{x} \sin x + e^{x} \cos x$ .

Worth noting

$- e^{x} \sin x$ multiplies the two derivatives, which is not the product rule.

Question 15

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Differentiate $\cot x$ .

\operatorname{cosec}^{2} x

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- \frac{2}{\sin 2x}

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- \operatorname{cosec}^{2} x

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2 \cot x \operatorname{cosec}^{2} x

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AnswerC · $- \operatorname{cosec}^{2} x$

$\frac{d}{dx}\left(\cot x\right) = - \operatorname{cosec}^{2} x$ , a standard derivative.
Keep the sign: the derivatives of the co-functions (cos, cot, cosec) are negative.

Worth noting

$\operatorname{cosec}^{2} x$ has the wrong sign.

Question 16

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What is the derivative of $\frac{\sin x}{x}$ ?

\frac{- x \cos x + \sin x}{x^{2}}

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\frac{x \cos x - \sin x}{x^{2}}

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\cos x

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\frac{x \cos x + \sin x}{x^{2}}

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AnswerB · $\frac{x \cos x - \sin x}{x^{2}}$

Quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ , with $u = \sin x$ and $v = x$ .
This simplifies to $\frac{x \cos x - \sin x}{x^{2}}$ .

Worth noting

$\frac{- x \cos x + \sin x}{x^{2}}$ has the numerator the wrong way round: it is $u'v - uv'$ , not $uv' - u'v$ .

Question 17

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What is $\frac{d}{dx}\left(\sqrt{x^{2} + 1}\right)$ ?

\frac{x}{\sqrt{x^{2} + 1}}

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\frac{1}{2 \sqrt{x^{2} + 1}}

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- \frac{x}{\sqrt{x^{2} + 1}}

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\frac{2 x}{\sqrt{x^{2} + 1}}

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AnswerA · $\frac{x}{\sqrt{x^{2} + 1}}$

Use the chain rule: differentiate the outer function, then multiply by the derivative of what is inside it.
$\frac{d}{dx}\left(\sqrt{x^{2} + 1}\right) = \frac{x}{\sqrt{x^{2} + 1}}$ .

Worth noting

$\frac{1}{2 \sqrt{x^{2} + 1}}$ differentiates the outside only and drops the inside's derivative.

Question 18

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Differentiate $x^{3} \log x$ with respect to $x$ .

3 x

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x^{2}

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3 x^{2} \log x - x^{2}

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3 x^{2} \log x + x^{2}

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AnswerD · $3 x^{2} \log x + x^{2}$

Product rule: $(uv)' = u'v + uv'$ , with $u = x^{3}$ and $v = \log x$ .
$(3 x^{2})(\log x) + (x^{3})(\frac{1}{x}) = 3 x^{2} \log x + x^{2}$ .

Worth noting

$3 x$ multiplies the two derivatives, which is not the product rule.

Question 19

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What is $\frac{d}{dx}\left(e^{x^{2}}\right)$ ?

e^{x^{2}}

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- 2 x e^{x^{2}}

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2 x e^{x^{2}}

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4 x e^{x^{2}}

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AnswerC · $2 x e^{x^{2}}$

Use the chain rule: differentiate the outer function, then multiply by the derivative of what is inside it.
$\frac{d}{dx}\left(e^{x^{2}}\right) = 2 x e^{x^{2}}$ .

Worth noting

$e^{x^{2}}$ differentiates the outside only and drops the inside's derivative.

Question 20

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What is $\frac{d}{dx}\left(\tan 2x\right)$ ?

\sec^{2} 2x

Correct answerYour answer

2 \sec^{2} 2x

Correct answerYour answer

- 2 \sec^{2} 2x

Correct answerYour answer

4 \sec^{2} 2x

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AnswerB · $2 \sec^{2} 2x$

Use the chain rule: differentiate the outer function, then multiply by the derivative of what is inside it.
$\frac{d}{dx}\left(\tan 2x\right) = 2 \sec^{2} 2x$ .

Worth noting

$\sec^{2} 2x$ differentiates the outside only and drops the inside's derivative.

Question 21

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$g(1) = 4$ , $g'(1) = 3$ , $h(1) = 2$ and $h'(1) = -1$ . If $f(x) = \frac{g(x)}{h(x)}$ , what is $f'(1)$ ?

- \frac{5}{2}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{5}{2}

Correct answerYour answer

5

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AnswerC · $\frac{5}{2}$

$f'(1) = \frac{g'(1)h(1) - g(1)h'(1)}{[h(1)]^{2}}$ .
$= \frac{(3)(2) - (4)(-1)}{4} = \frac{10}{4} = \frac{5}{2}$ .

Worth noting

$-\frac{5}{2}$ puts the numerator the wrong way round, $gh' - g'h$ .

Question 22

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What is $\frac{d}{dx}\left(\sin x\cos x\right)$ ?

\cos 2x

Correct answerYour answer

- \sin x \cos x

Correct answerYour answer

1

Correct answerYour answer

\sin 2x

Correct answerYour answer

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AnswerA · $\cos 2x$

Product rule: $\cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^{2} x - \sin^{2} x$ .
$\cos^{2} x - \sin^{2} x = \cos 2x$ . (Or first write $\sin x\cos x = \frac{1}{2}\sin 2x$ .)

Worth noting

$-\sin x\cos x$ multiplies the two derivatives, $\cos x$ and $-\sin x$ , which is not the product rule.

Question 23

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Which of the following statements are true?

I. $\frac{d}{dx}\left(5^{x}\right) = x \cdot 5^{x - 1}$ .

II. $\frac{d}{dx}\left(\log_{10} x\right) = \frac{1}{x\log 10}$ .

III. $\frac{d}{dx}\left(e^{-3x}\right) = -3e^{-3x}$ .

IV. $\frac{d}{dx}\left(\log 7x\right) = \frac{7}{x}$ .

II, III and IV onlyCorrect answerYour answerI and III onlyCorrect answerYour answerIII onlyCorrect answerYour answerII and III onlyCorrect answerYour answer

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AnswerD · II and III only

II: $\log_{10} x = \frac{\log x}{\log 10}$ , whose derivative is $\frac{1}{x\log 10}$ . III: the chain rule gives the factor $-3$ .
I is false: the power rule is for a variable base; $\frac{d}{dx}5^{x} = 5^{x}\log 5$ . IV is false: $\log 7x = \log 7 + \log x$ , so its derivative is $\frac{1}{x}$ .

Worth noting

IV is the common slip: the 7 inside the logarithm cancels, $\frac{7}{7x} = \frac{1}{x}$ .

Question 24

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What is $\frac{d}{dx}\left(2^{3x}\right)$ ?

3 \cdot 2^{3 x} \log 2

Correct answerYour answer

2^{3 x} \log 2

Correct answerYour answer

3 \cdot 2^{3 x - 1} x

Correct answerYour answer

3 \cdot 2^{3 x}

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AnswerA · $3 \cdot 2^{3 x} \log 2$

$\frac{d}{dx}a^{u} = a^{u}\log a \cdot \frac{du}{dx}$ , with $u = 3x$ .
$2^{3x}\log 2 \cdot 3 = 3 \cdot 2^{3x}\log 2$ .

Worth noting

$2^{3x}\log 2$ forgets the factor 3 from the inside, $3x$ .

Question 25

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If $f(x) = x\sqrt{x + 3}$ , what is $f'(1)$ ?

\frac{1}{4}

Correct answerYour answer

2

Correct answerYour answer

\frac{9}{4}

Correct answerYour answer

\frac{5}{2}

Correct answerYour answer

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AnswerC · $\frac{9}{4}$

Product rule: $f'(x) = \sqrt{x + 3} + x \cdot \frac{1}{2\sqrt{x + 3}}$ .
At $x = 1$ : $\sqrt{4} + \frac{1}{2 \times 2} = 2 + \frac{1}{4} = \frac{9}{4}$ .

Worth noting

$\frac{1}{4}$ keeps only $x$ times the derivative of the root and drops the term $\sqrt{x + 3}$ .

Question 26

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Match each function in List I with its derivative in List II.

List I	List II
A. $\tan x$	I. $\sec x\tan x$
B. $\sec x$	II. $-\operatorname{cosec} x\cot x$
C. $\cot x$	III. $\sec^{2} x$
D. $\operatorname{cosec} x$	IV. $-\operatorname{cosec}^{2} x$

A-I, B-III, C-II, D-IVCorrect answerYour answerA-III, B-I, C-IV, D-IICorrect answerYour answerA-III, B-I, C-II, D-IVCorrect answerYour answerA-I, B-III, C-IV, D-IICorrect answerYour answer

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AnswerB · A-III, B-I, C-IV, D-II

$\frac{d}{dx}\tan x = \sec^{2} x$ and $\frac{d}{dx}\sec x = \sec x\tan x$ .
The co-functions carry a minus sign: $\frac{d}{dx}\cot x = -\operatorname{cosec}^{2} x$ and $\frac{d}{dx}\operatorname{cosec} x = -\operatorname{cosec} x\cot x$ .

Question 27

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What is $\frac{d}{dx}\left(\sin^2 3x\right)$ ?

2 \sin 3x

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\sin 6x

Correct answerYour answer

6 \cos^{2} 3x

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3 \sin 6x

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AnswerD · $3 \sin 6x$

Three layers: the square, then the sine, then $3x$ . Differentiate each and multiply.
$2\sin 3x \cdot \cos 3x \cdot 3 = 6\sin 3x\cos 3x = 3\sin 6x$ .

Worth noting

$\sin 6x$ is $2\sin 3x\cos 3x$ : the factor 3 from the inner $3x$ is missing.

Question 28

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What is the derivative of $\frac{x}{x^{2} + 1}$ ?

\frac{x^{2} - 1}{\left(x^{2} + 1\right)^{2}}

Correct answerYour answer

\frac{1}{2 x}

Correct answerYour answer

\frac{3 x^{2} + 1}{\left(x^{2} + 1\right)^{2}}

Correct answerYour answer

\frac{1 - x^{2}}{\left(x^{2} + 1\right)^{2}}

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AnswerD · $\frac{1 - x^{2}}{\left(x^{2} + 1\right)^{2}}$

Quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ , with $u = x$ and $v = x^{2} + 1$ .
This simplifies to $\frac{1 - x^{2}}{\left(x^{2} + 1\right)^{2}}$ .

Worth noting

$\frac{x^{2} - 1}{\left(x^{2} + 1\right)^{2}}$ has the numerator the wrong way round: it is $u'v - uv'$ , not $uv' - u'v$ .

Implicit functions

18 questions

Question 29

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Find $\frac{dy}{dx}$ at $(3, 4)$ for the curve $x^2 + y^2 = 25$ .

- \frac{4}{3}

Correct answerYour answer

- \frac{3}{4}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{3}{4}

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AnswerB · $- \frac{3}{4}$

Differentiate every term with respect to $x$ , treating $y$ as a function of $x$ , so $y$ terms pick up $\frac{dy}{dx}$ .
Solving for it: $\frac{dy}{dx} = - \frac{x}{y}$ .
At $(3, 4)$ this is $- \frac{3}{4}$ .

Worth noting

$- \frac{4}{3}$ is $\frac{dx}{dy}$ , the reciprocal.

Question 30

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Find $\frac{dy}{dx}$ at $(3, 4)$ for the curve $xy = 12$ .

- \frac{4}{3}

Correct answerYour answer

- \frac{3}{4}

Correct answerYour answer

- \frac{1}{3}

Correct answerYour answer

\frac{4}{3}

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AnswerA · $- \frac{4}{3}$

Differentiate every term with respect to $x$ , treating $y$ as a function of $x$ , so $y$ terms pick up $\frac{dy}{dx}$ .
Solving for it: $\frac{dy}{dx} = - \frac{y}{x}$ .
At $(3, 4)$ this is $- \frac{4}{3}$ .

Worth noting

$- \frac{3}{4}$ is $\frac{dx}{dy}$ , the reciprocal.

Question 31

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Find $\frac{dy}{dx}$ at $(2, 2)$ for the curve $x^2 + 4y^2 = 20$ .

-4

Correct answerYour answer

-1

Correct answerYour answer

- \frac{1}{4}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

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AnswerC · $- \frac{1}{4}$

Differentiate term by term: $2x + 8y\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -\frac{x}{4y}$ .
At $(2, 2)$ : $-\frac{2}{8} = -\frac{1}{4}$ .

Worth noting

$-1$ is $-\frac{x}{y}$ , the circle's slope: the 4 in front of $y^2$ was lost.

Question 32

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If $x^2 + 2xy + 3y^2 = 1$ , what is $\frac{dy}{dx}$ ?

\frac{- x - 3 y}{x + y}

Correct answerYour answer

- \frac{x}{3 y}

Correct answerYour answer

\frac{x + y}{x + 3 y}

Correct answerYour answer

\frac{- x - y}{x + 3 y}

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AnswerD · $\frac{- x - y}{x + 3 y}$

$2x + 2\left(y + x\frac{dy}{dx}\right) + 6y\frac{dy}{dx} = 0$ : the term $2xy$ needs the product rule.
$(2x + 6y)\frac{dy}{dx} = -(2x + 2y)$ , so $\frac{dy}{dx} = -\frac{x + y}{x + 3y}$ .

Worth noting

$-\frac{x}{3y}$ differentiates $2xy$ as if $y$ were a constant, dropping the product rule.

Question 33

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If $y + \cos y = x^2$ , what is $\frac{dy}{dx}$ ?

\frac{2 x}{1 - \sin y}

Correct answerYour answer

\frac{2 x}{\sin y + 1}

Correct answerYour answer

2 x

Correct answerYour answer

2 x \left(1 - \sin y\right)

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AnswerA · $\frac{2 x}{1 - \sin y}$

Differentiate both sides: $\frac{dy}{dx} - \sin y\frac{dy}{dx} = 2x$ .
$(1 - \sin y)\frac{dy}{dx} = 2x$ , so $\frac{dy}{dx} = \frac{2x}{1 - \sin y}$ .

Worth noting

$\frac{2x}{1 + \sin y}$ takes the derivative of $\cos y$ as $+\sin y$ ; it is $-\sin y \cdot \frac{dy}{dx}$ .

Question 34

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If $e^y = x$ , what is $\frac{dy}{dx}$ ?

e^{y}

Correct answerYour answer

x

Correct answerYour answer

\frac{1}{x}

Correct answerYour answer

- \frac{1}{x}

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AnswerC · $\frac{1}{x}$

Differentiate both sides: $e^y\frac{dy}{dx} = 1$ .
$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$ .

Worth noting

$e^y$ is $\frac{dx}{dy}$ ; the question asks for its reciprocal.

Question 35

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If $x = \sin y$ , what is $\frac{dy}{dx}$ in terms of $x$ ?

\sqrt{1 - x^{2}}

Correct answerYour answer

- \frac{1}{\sqrt{1 - x^{2}}}

Correct answerYour answer

\frac{1}{\cos x}

Correct answerYour answer

\frac{1}{\sqrt{1 - x^{2}}}

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AnswerD · $\frac{1}{\sqrt{1 - x^{2}}}$

Differentiate with respect to $x$ : $1 = \cos y\frac{dy}{dx}$ , so $\frac{dy}{dx} = \frac{1}{\cos y}$ .
$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ (taking $\cos y > 0$ ), so $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ .

Worth noting

$\frac{1}{\cos x}$ puts $x$ where $y$ belongs: the derivative is $\frac{1}{\cos y}$ , rewritten with $\sin y = x$ .

Question 36

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Which of the following statements are true?

I. $\frac{d}{dx}\left(y^2\right) = 2y\frac{dy}{dx}$ .

II. $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ .

III. $\frac{d}{dx}(\sin y) = \cos y$ .

IV. For $x^2 + y^2 = 16$ , $\frac{dy}{dx} = -\frac{x}{y}$ .

I, II and IV onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerII and IV onlyCorrect answerYour answer

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AnswerA · I, II and IV only

Every function of $y$ picks up a factor $\frac{dy}{dx}$ when differentiated with respect to $x$ : that makes I true and III false, since $\frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx}$ .
II is the product rule. IV: $2x + 2y\frac{dy}{dx} = 0$ gives $-\frac{x}{y}$ , so it is true.

Question 37

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Find $\frac{dy}{dx}$ at $(1, 2)$ for the curve $x^2 + xy + y^2 = 7$ .

- \frac{8}{5}

Correct answerYour answer

- \frac{5}{4}

Correct answerYour answer

- \frac{4}{5}

Correct answerYour answer

\frac{4}{5}

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AnswerC · $- \frac{4}{5}$

Differentiate every term with respect to $x$ , treating $y$ as a function of $x$ , so $y$ terms pick up $\frac{dy}{dx}$ .
Solving for it: $\frac{dy}{dx} = \frac{- 2 x - y}{x + 2 y}$ .
At $(1, 2)$ this is $- \frac{4}{5}$ .

Worth noting

$- \frac{5}{4}$ is $\frac{dx}{dy}$ , the reciprocal.

Question 38

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Find $\frac{dy}{dx}$ at $(5, 3)$ for the curve $x^2 - y^2 = 16$ .

- \frac{5}{3}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

\frac{5}{3}

Correct answerYour answer

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AnswerD · $\frac{5}{3}$

Differentiate every term with respect to $x$ , treating $y$ as a function of $x$ , so $y$ terms pick up $\frac{dy}{dx}$ .
Solving for it: $\frac{dy}{dx} = \frac{x}{y}$ .
At $(5, 3)$ this is $\frac{5}{3}$ .

Worth noting

$\frac{3}{5}$ is $\frac{dx}{dy}$ , the reciprocal.

Question 39

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Assertion (A): For $x^2 + y^2 = 4$ , $\frac{dy}{dx} = -\frac{x}{y}$ .

Reason (R): Differentiating $y^2$ with respect to $x$ gives $2y\frac{dy}{dx}$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerA · Both A and R are true, and R is the correct explanation of A.

R is the chain rule, with $y$ a function of $x$ , so it is true.
Using it, $2x + 2y\frac{dy}{dx} = 0$ and $\frac{dy}{dx} = -\frac{x}{y}$ : A is true, and R is the step that gives it.

Question 40

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Find $\frac{dy}{dx}$ at $(1, 2)$ for the curve $x^2y + xy^2 = 6$ .

- \frac{8}{5}

Correct answerYour answer

- \frac{5}{8}

Correct answerYour answer

- \frac{3}{5}

Correct answerYour answer

\frac{8}{5}

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AnswerA · $- \frac{8}{5}$

Product rule on each term: $2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$ .
At $(1, 2)$ : $4 + \frac{dy}{dx} + 4 + 4\frac{dy}{dx} = 0$ , so $5\frac{dy}{dx} = -8$ and $\frac{dy}{dx} = -\frac{8}{5}$ .

Worth noting

$-\frac{5}{8}$ divides the wrong way: the terms with $\frac{dy}{dx}$ add to $5\frac{dy}{dx}$ , and the rest to 8.

Question 41

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If $y = x + \log y$ , what is $\frac{dy}{dx}$ ?

\frac{y - 1}{y}

Correct answerYour answer

\frac{y}{1 - y}

Correct answerYour answer

1 + \frac{1}{y}

Correct answerYour answer

\frac{y}{y - 1}

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AnswerD · $\frac{y}{y - 1}$

Differentiate both sides: $\frac{dy}{dx} = 1 + \frac{1}{y}\frac{dy}{dx}$ .
$\frac{dy}{dx}\left(1 - \frac{1}{y}\right) = 1$ , so $\frac{dy}{dx} = \frac{y}{y - 1}$ .

Worth noting

$1 + \frac{1}{y}$ differentiates $\log y$ as $\frac{1}{y}$ and forgets the factor $\frac{dy}{dx}$ .

Question 42

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Read the two statements and choose the correct option.

I. For $y^2 = 4x$ , $\frac{dy}{dx} = \frac{2}{y}$ .

II. For $x^2 - y^2 = 1$ , $\frac{dy}{dx} = -\frac{x}{y}$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $2y\frac{dy}{dx} = 4$ , so $\frac{dy}{dx} = \frac{2}{y}$ . True.
II: $2x - 2y\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = \frac{x}{y}$ , with no minus sign. False.

Worth noting

II copies the circle's slope; the minus in front of $y^2$ changes the sign.

Question 43

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If $e^{x} + e^{y} = 1$ , what is $\frac{dy}{dx}$ ?

- e^{x - y}

Correct answerYour answer

e^{x - y}

Correct answerYour answer

- e^{- x + y}

Correct answerYour answer

- e^{x}

Correct answerYour answer

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AnswerA · $- e^{x - y}$

$e^{x} + e^{y}\frac{dy}{dx} = 0$ .
$\frac{dy}{dx} = -\frac{e^{x}}{e^{y}} = -e^{x - y}$ .

Worth noting

$-e^{y - x}$ divides the wrong way round: it is $e^{x}$ over $e^{y}$ .

Question 44

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Find $\frac{dy}{dx}$ at $(4, 1)$ for the curve $\sqrt{x} + \sqrt{y} = 3$ .

-2

Correct answerYour answer

- \frac{1}{2}

Correct answerYour answer

- \frac{1}{4}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerB · $- \frac{1}{2}$

$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$ .
At $(4, 1)$ : $-\frac{1}{2}$ .

Worth noting

$-2$ is $-\frac{\sqrt{x}}{\sqrt{y}}$ , the ratio turned upside down.

Question 45

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If $xy = e^{x - y}$ , what is $\frac{dy}{dx}$ ?

\frac{x \left(y + 1\right)}{y \left(x - 1\right)}

Correct answerYour answer

\frac{y \left(x - 1\right)}{x \left(y + 1\right)}

Correct answerYour answer

\frac{y \left(x + 1\right)}{x \left(y - 1\right)}

Correct answerYour answer

\frac{x - 1}{y + 1}

Correct answerYour answer

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AnswerB · $\frac{y \left(x - 1\right)}{x \left(y + 1\right)}$

Take logarithms: $\log x + \log y = x - y$ .
Differentiate: $\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx}$ , so $\frac{dy}{dx}\left(\frac{1}{y} + 1\right) = 1 - \frac{1}{x}$ .
$\frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)}$ .

Worth noting

The flipped fraction is $\frac{dx}{dy}$ .

Question 46

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Find $\frac{dy}{dx}$ at $(2, 4)$ for the curve $x^3 + y^3 = 9xy$ .

- \frac{4}{5}

Correct answerYour answer

\frac{4}{5}

Correct answerYour answer

\frac{5}{4}

Correct answerYour answer

\frac{9}{5}

Correct answerYour answer

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AnswerB · $\frac{4}{5}$

Differentiate every term with respect to $x$ , treating $y$ as a function of $x$ , so $y$ terms pick up $\frac{dy}{dx}$ .
Solving for it: $\frac{dy}{dx} = \frac{x^{2} - 3 y}{3 x - y^{2}}$ .
At $(2, 4)$ this is $\frac{4}{5}$ .

Worth noting

$\frac{5}{4}$ is $\frac{dx}{dy}$ , the reciprocal.

Parametric functions

15 questions

Question 47

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If $x = 2 t^{2}$ and $y = 4 t$ , what is $\frac{dy}{dx}$ ?

t

Correct answerYour answer

4

Correct answerYour answer

\frac{1}{t}

Correct answerYour answer

- \frac{1}{t}

Correct answerYour answer

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AnswerC · $\frac{1}{t}$

$\frac{dx}{dt} = 4 t$ and $\frac{dy}{dt} = 4$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{t}$ .

Worth noting

Dividing the other way round gives $\frac{dx}{dy}$ .

Question 48

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A curve is given by $x = t^{2} + 1$ , $y = t^{3}$ . What is its slope $\frac{dy}{dx}$ at $t = 2$ ?

\frac{1}{3}

Correct answerYour answer

3

Correct answerYour answer

4

Correct answerYour answer

6

Correct answerYour answer

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AnswerB · $3$

$\frac{dx}{dt} = 2 t$ and $\frac{dy}{dt} = 3 t^{2}$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 t}{2}$ .
At $t = 2$ the slope is $3$ .

Worth noting

Dividing the other way round gives $\frac{dx}{dy}$ .

Question 49

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A curve is given by $x = 3t^{2} + 1$ , $y = t^{3} - 2t$ . What is its slope $\frac{dy}{dx}$ at $t = 1$ ?

- \frac{5}{6}

Correct answerYour answer

- \frac{1}{3}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

6

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AnswerC · $\frac{1}{6}$

$\frac{dx}{dt} = 6t = 6$ and $\frac{dy}{dt} = 3t^{2} - 2 = 1$ at $t = 1$ .
$\frac{dy}{dx} = \frac{1}{6}$ .

Worth noting

6 is $\frac{dx}{dt} \div \frac{dy}{dt}$ , the slope turned upside down.

Question 50

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If $x = \cos^3 t$ and $y = \sin^3 t$ , what is $\frac{dy}{dx}$ ?

\tan t

Correct answerYour answer

- \cot t

Correct answerYour answer

- \tan^{2} t

Correct answerYour answer

- \tan t

Correct answerYour answer

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AnswerD · $- \tan t$

$\frac{dx}{dt} = -3\cos^2 t\sin t$ and $\frac{dy}{dt} = 3\sin^2 t\cos t$ .
$\frac{dy}{dx} = \frac{3\sin^2 t\cos t}{-3\cos^2 t\sin t} = -\frac{\sin t}{\cos t} = -\tan t$ .

Worth noting

$\tan t$ loses the minus sign that comes from differentiating $\cos t$ .

Question 51

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Read the two statements and choose the correct option.

I. If $x = f(t)$ and $y = g(t)$ , then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ wherever $\frac{dx}{dt} \ne 0$ .

II. If $x = f(t)$ and $y = g(t)$ , then $\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2}$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I is the chain rule $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ , divided through. It is true.
II is false. For $x = t^2$ , $y = t^3$ : $\frac{dy}{dx} = \frac{3t}{2}$ , so $\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$ , while the ratio in II is $\frac{6t}{2} = 3t$ .

Question 52

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A curve is given by $x = 3 \cos t$ , $y = 2 \sin t$ . What is its slope $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$ ?

- \frac{3 \sqrt{3}}{2}

Correct answerYour answer

- \frac{4 \sqrt{3}}{9}

Correct answerYour answer

- \frac{2 \sqrt{3}}{9}

Correct answerYour answer

\frac{2 \sqrt{3}}{9}

Correct answerYour answer

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AnswerC · $- \frac{2 \sqrt{3}}{9}$

$\frac{dx}{dt} = - 3 \sin t$ and $\frac{dy}{dt} = 2 \cos t$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{2}{3 \tan t}$ .
At $t = \frac{\pi}{3}$ the slope is $- \frac{2 \sqrt{3}}{9}$ .

Worth noting

Dividing the other way round gives $\frac{dx}{dy}$ .

Question 53

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A curve is given by $x = e^{t}$ , $y = e^{- t}$ . What is its slope $\frac{dy}{dx}$ at $t = \log 2$ ?

-4

Correct answerYour answer

- \frac{5}{4}

Correct answerYour answer

- \frac{1}{2}

Correct answerYour answer

- \frac{1}{4}

Correct answerYour answer

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AnswerD · $- \frac{1}{4}$

$\frac{dx}{dt} = e^{t}$ and $\frac{dy}{dt} = - e^{- t}$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - e^{- 2 t}$ .
At $t = \log 2$ the slope is $- \frac{1}{4}$ .

Worth noting

Dividing the other way round gives $\frac{dx}{dy}$ .

Question 54

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A curve is given by $x = \log t$ , $y = t^{2}$ . What is its slope $\frac{dy}{dx}$ at $t = 1$ ?

-2

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

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AnswerC · $2$

$\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = 2 t$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 2 t^{2}$ .
At $t = 1$ the slope is $2$ .

Worth noting

Dividing the other way round gives $\frac{dx}{dy}$ .

Question 55

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If $x = \sec t$ and $y = \tan t$ , what is $\frac{dy}{dx}$ ?

\sin t

Correct answerYour answer

- \operatorname{cosec} t

Correct answerYour answer

\operatorname{cosec} t

Correct answerYour answer

\cot t

Correct answerYour answer

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AnswerC · $\operatorname{cosec} t$

$\frac{dx}{dt} = \sec t\tan t$ and $\frac{dy}{dt} = \sec^{2} t$ .
$\frac{dy}{dx} = \frac{\sec^{2} t}{\sec t\tan t} = \frac{\sec t}{\tan t} = \frac{1}{\sin t} = \operatorname{cosec} t$ .

Worth noting

$\sin t$ is $\frac{dx}{dt} \div \frac{dy}{dt}$ , the ratio upside down.

Question 56

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For the curve $x = t^{2} + 1$ , $y = t^{3} - 3t$ , at which values of $t$ is the tangent horizontal?

t = 0

onlyCorrect answerYour answer

t = 1

onlyCorrect answerYour answer

t = \sqrt{3}

and

t = -\sqrt{3}

Correct answerYour answer

t = 1

and

t = -1

Correct answerYour answer

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AnswerD · $t = 1$ and $t = -1$

The tangent is horizontal where $\frac{dy}{dx} = 0$ : $\frac{dy}{dt} = 0$ while $\frac{dx}{dt} \ne 0$ .
$3t^{2} - 3 = 0$ gives $t = \pm 1$ , and there $\frac{dx}{dt} = 2t \ne 0$ .

Worth noting

At $t = 0$ it is $\frac{dx}{dt}$ that is 0: the tangent there is vertical, not horizontal. $t = \pm\sqrt{3}$ is where $y = 0$ .

Question 57

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What is the derivative of $x^{3}$ with respect to $x^{2}$ ?

\frac{2}{3 x}

Correct answerYour answer

\frac{3 x}{2}

Correct answerYour answer

x

Correct answerYour answer

3 x^{2}

Correct answerYour answer

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AnswerB · $\frac{3 x}{2}$

Let $u = x^{3}$ and $v = x^{2}$ ; then $\frac{du}{dv} = \frac{du/dx}{dv/dx}$ , with $x$ as the parameter.
$\frac{3x^{2}}{2x} = \frac{3x}{2}$ .

Worth noting

$3x^{2}$ is the derivative with respect to $x$ ; it still has to be divided by $\frac{d}{dx}(x^{2}) = 2x$ .

Question 58

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What is the derivative of $\sin x$ with respect to $\cos x$ ?

\cot x

Correct answerYour answer

- \tan x

Correct answerYour answer

\tan x

Correct answerYour answer

- \cot x

Correct answerYour answer

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AnswerD · $- \cot x$

$\frac{d(\sin x)}{d(\cos x)} = \frac{\cos x}{-\sin x}$ .
$= -\cot x$ .

Worth noting

$\cot x$ forgets that the derivative of $\cos x$ is $-\sin x$ .

Question 59

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If $x = 2(t - \sin t)$ and $y = 2(1 - \cos t)$ , what is $\frac{dy}{dx}$ ?

\cot\frac{t}{2}

Correct answerYour answer

\tan\frac{t}{2}

Correct answerYour answer

\sin t

Correct answerYour answer

2\cot\frac{t}{2}

Correct answerYour answer

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AnswerA · $\cot\frac{t}{2}$

$\frac{dx}{dt} = 2(1 - \cos t)$ and $\frac{dy}{dt} = 2\sin t$ , so $\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}$ .
With $\sin t = 2\sin\frac{t}{2}\cos\frac{t}{2}$ and $1 - \cos t = 2\sin^2\frac{t}{2}$ , this is $\cot\frac{t}{2}$ .

Worth noting

$\tan\frac{t}{2}$ is $\frac{dx}{dy}$ , the division the other way round.

Question 60

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A curve is given by $x = t - \frac{1}{t}$ , $y = t + \frac{1}{t}$ . What is its slope $\frac{dy}{dx}$ at $t = 2$ ?

- \frac{3}{5}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{8}{5}

Correct answerYour answer

\frac{5}{3}

Correct answerYour answer

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AnswerB · $\frac{3}{5}$

$\frac{dx}{dt} = 1 + \frac{1}{t^{2}}$ and $\frac{dy}{dt} = 1 - \frac{1}{t^{2}}$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t^{2} - 1}{t^{2} + 1}$ .
At $t = 2$ the slope is $\frac{3}{5}$ .

Worth noting

Dividing the other way round gives $\frac{dx}{dy}$ .

Question 61

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If $x = a(\cos t + t\sin t)$ and $y = a(\sin t - t\cos t)$ , what is $\frac{dy}{dx}$ ?

\cot t

Correct answerYour answer

\tan t

Correct answerYour answer

t \tan t

Correct answerYour answer

- \tan t

Correct answerYour answer

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AnswerB · $\tan t$

$\frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t$ .
$\frac{dy}{dt} = a(\cos t - \cos t + t\sin t) = at\sin t$ , so $\frac{dy}{dx} = \frac{at\sin t}{at\cos t} = \tan t$ .

Worth noting

$\cot t$ divides $\frac{dx}{dt}$ by $\frac{dy}{dt}$ .

Second-order derivatives

21 questions

Question 62

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If $y = x^{4} - 3 x^{2} + 7 x$ , what is $\frac{d^2y}{dx^2}$ ?

4 x^{3} - 6 x + 7

Correct answerYour answer

12 x^{2} - 6

Correct answerYour answer

6 - 12 x^{2}

Correct answerYour answer

\left(4 x^{3} - 6 x + 7\right)^{2}

Correct answerYour answer

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AnswerB · $12 x^{2} - 6$

First derivative: $\frac{dy}{dx} = 4 x^{3} - 6 x + 7$ .
Differentiate again: $\frac{d^2y}{dx^2} = 12 x^{2} - 6$ .

Worth noting

$4 x^{3} - 6 x + 7$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 63

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If $y = \sin 2x$ , what is $\frac{d^2y}{dx^2}$ ?

2 \cos 2x

Correct answerYour answer

4 \sin 2x

Correct answerYour answer

4 \cos^{2} 2x

Correct answerYour answer

- 4 \sin 2x

Correct answerYour answer

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AnswerD · $- 4 \sin 2x$

First derivative: $\frac{dy}{dx} = 2 \cos 2x$ .
Differentiate again: $\frac{d^2y}{dx^2} = - 4 \sin 2x$ .

Worth noting

$2 \cos 2x$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 64

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For $f(x) = x^{3} + 2 x^{2} - x$ , what is $f''(1)$ ?

6

Correct answerYour answer

10

Correct answerYour answer

12

Correct answerYour answer

20

Correct answerYour answer

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AnswerB · $10$

$f'(x) = 3 x^{2} + 4 x - 1$ and $f''(x) = 6 x + 4$ .
So $f''(1) = 10$ .

Worth noting

$6$ is $f'(1)$ , one derivative short.

Question 65

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For $f(x) = e^{3 x}$ , what is $f''(0)$ ?

-9

Correct answerYour answer

3

Correct answerYour answer

8

Correct answerYour answer

9

Correct answerYour answer

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AnswerD · $9$

$f'(x) = 3 e^{3 x}$ and $f''(x) = 9 e^{3 x}$ .
So $f''(0) = 9$ .

Worth noting

$3$ is $f'(0)$ , one derivative short.

Question 66

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If $y = \frac{1}{x}$ , what is $\frac{d^2y}{dx^2}$ ?

\frac{2}{x^{3}}

Correct answerYour answer

- \frac{2}{x^{3}}

Correct answerYour answer

- \frac{1}{x^{2}}

Correct answerYour answer

\frac{1}{x^{3}}

Correct answerYour answer

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AnswerA · $\frac{2}{x^{3}}$

$y = x^{-1}$ , so $\frac{dy}{dx} = -x^{-2}$ .
$\frac{d^2y}{dx^2} = 2x^{-3} = \frac{2}{x^{3}}$ .

Worth noting

$-\frac{2}{x^{3}}$ loses a sign: the power $-2$ times the $-1$ already there is $+2$ .

Question 67

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For $f(x) = \left(x^{2} - 1\right)^{2}$ , what is $f''(2)$ ?

24

Correct answerYour answer

40

Correct answerYour answer

44

Correct answerYour answer

48

Correct answerYour answer

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AnswerC · $44$

Expand: $f(x) = x^{4} - 2x^{2} + 1$ , so $f'(x) = 4x^{3} - 4x$ and $f''(x) = 12x^{2} - 4$ .
$f''(2) = 48 - 4 = 44$ .

Worth noting

24 is $f'(2)$ , the first derivative.

Question 68

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If $y = \log x$ , what is $\frac{d^2y}{dx^2}$ ?

\frac{1}{x}

Correct answerYour answer

\frac{1}{x^{2}}

Correct answerYour answer

- \frac{1}{x^{2}}

Correct answerYour answer

\frac{x - 1}{x^{2}}

Correct answerYour answer

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AnswerC · $- \frac{1}{x^{2}}$

First derivative: $\frac{dy}{dx} = \frac{1}{x}$ .
Differentiate again: $\frac{d^2y}{dx^2} = - \frac{1}{x^{2}}$ .

Worth noting

$\frac{1}{x}$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 69

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If $y = x e^{x}$ , what is $\frac{d^2y}{dx^2}$ ?

\left(x + 1\right) e^{x}

Correct answerYour answer

\left(x + 2\right) e^{x}

Correct answerYour answer

- \left(x + 2\right) e^{x}

Correct answerYour answer

\left(x + 1\right)^{2} e^{2 x}

Correct answerYour answer

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AnswerB · $\left(x + 2\right) e^{x}$

First derivative: $\frac{dy}{dx} = \left(x + 1\right) e^{x}$ .
Differentiate again: $\frac{d^2y}{dx^2} = \left(x + 2\right) e^{x}$ .

Worth noting

$\left(x + 1\right) e^{x}$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 70

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For $f(x) = x^{3} - 6 x^{2} + 9 x$ , at what value of $x$ is $f''(x) = 0$ ?

0

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

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AnswerC · $2$

$f'(x) = 3x^2 - 12x + 9$ and $f''(x) = 6x - 12$ .
$6x - 12 = 0$ gives $x = 2$ .

Worth noting

1 and 3 are where $f'(x) = 0$ , the first derivative.

Question 71

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Assertion (A): If $y = x^3$ , then $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$ .

Reason (R): $\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ .

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AnswerD · A is false, but R is true.

$\frac{dy}{dx} = 3x^2$ , so $\frac{d^2y}{dx^2} = 6x$ , while $\left(\frac{dy}{dx}\right)^2 = 9x^4$ : A is false.
R is the definition of the second derivative, so it is true.

Question 72

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For $f(x) = \log{\left(x + 1 \right)}$ , what is $f''(1)$ ?

- \frac{1}{4}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{7}{4}

Correct answerYour answer

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AnswerA · $- \frac{1}{4}$

$f'(x) = \frac{1}{x + 1}$ and $f''(x) = - \frac{1}{\left(x + 1\right)^{2}}$ .
So $f''(1) = - \frac{1}{4}$ .

Worth noting

$\frac{1}{2}$ is $f'(1)$ , one derivative short.

Question 73

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If $y = e^{x} \sin x$ , what is $\frac{d^2y}{dx^2}$ ?

2 e^{x} \cos x

Correct answerYour answer

\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}

Correct answerYour answer

- 2 e^{x} \cos x

Correct answerYour answer

2 e^{2 x} \sin^{2}{\left(x + \frac{\pi}{4} \right)}

Correct answerYour answer

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AnswerA · $2 e^{x} \cos x$

First derivative: $\frac{dy}{dx} = \sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}$ .
Differentiate again: $\frac{d^2y}{dx^2} = 2 e^{x} \cos x$ .

Worth noting

$\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 74

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If $y = \tan x$ , what is $\frac{d^2y}{dx^2}$ ?

\sec^{2} x

Correct answerYour answer

2 \tan x \sec x

Correct answerYour answer

2 \tan x \sec^{2} x

Correct answerYour answer

2 \sec^{2} x

Correct answerYour answer

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AnswerC · $2 \tan x \sec^{2} x$

$\frac{dy}{dx} = \sec^{2} x = (\sec x)^{2}$ .
Chain rule: $2\sec x \cdot \sec x\tan x = 2\sec^{2} x\tan x$ .

Worth noting

$2\sec x\tan x$ differentiates the square but forgets to multiply by $\sec x$ once more, from $\frac{d}{dx}\sec x = \sec x\tan x$ .

Question 75

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If $y = 2^{x}$ , what is $\frac{d^2y}{dx^2}$ ?

2^{x} \log 2

Correct answerYour answer

2^{x} \left(\log 2\right)^{2}

Correct answerYour answer

2^{x - 2} x \left(x - 1\right)

Correct answerYour answer

2 \cdot 2^{x} \log 2

Correct answerYour answer

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AnswerB · $2^{x} \left(\log 2\right)^{2}$

$\frac{dy}{dx} = 2^{x}\log 2$ , and $\log 2$ is a constant.
$\frac{d^2y}{dx^2} = \log 2 \cdot 2^{x}\log 2 = 2^{x}(\log 2)^{2}$ .

Worth noting

$2^{x}\log 2$ is the first derivative; differentiating again brings out another $\log 2$ .

Question 76

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If $y = \sqrt{x}$ , what is $\frac{d^2y}{dx^2}$ ?

- \frac{1}{4 x^{\frac{3}{2}}}

Correct answerYour answer

\frac{1}{4 x^{\frac{3}{2}}}

Correct answerYour answer

- \frac{1}{2 x^{\frac{3}{2}}}

Correct answerYour answer

\frac{1}{2 \sqrt{x}}

Correct answerYour answer

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AnswerA · $- \frac{1}{4 x^{\frac{3}{2}}}$

$y = x^{1/2}$ , so $\frac{dy}{dx} = \frac{1}{2}x^{-1/2}$ .
$\frac{d^2y}{dx^2} = \frac{1}{2} \cdot \left(-\frac{1}{2}\right)x^{-3/2} = -\frac{1}{4x^{3/2}}$ .

Worth noting

$\frac{1}{4x^{3/2}}$ drops the minus sign that the power $-\frac{1}{2}$ brings down.

Question 77

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If $y = \cos{\left(x^{2} \right)}$ , what is $\frac{d^2y}{dx^2}$ ?

- 2 x \sin{\left(x^{2} \right)}

Correct answerYour answer

4 x^{2} \cos{\left(x^{2} \right)} + 2 \sin{\left(x^{2} \right)}

Correct answerYour answer

- 4 x^{2} \cos{\left(x^{2} \right)} - 2 \sin{\left(x^{2} \right)}

Correct answerYour answer

4 x^{2} \sin^{2}{\left(x^{2} \right)}

Correct answerYour answer

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AnswerC · $- 4 x^{2} \cos{\left(x^{2} \right)} - 2 \sin{\left(x^{2} \right)}$

First derivative: $\frac{dy}{dx} = - 2 x \sin{\left(x^{2} \right)}$ .
Differentiate again: $\frac{d^2y}{dx^2} = - 4 x^{2} \cos{\left(x^{2} \right)} - 2 \sin{\left(x^{2} \right)}$ .

Worth noting

$- 2 x \sin{\left(x^{2} \right)}$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 78

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If $x = t^2$ and $y = t^3$ , what is $\frac{d^2y}{dx^2}$ at $t = 1$ ?

\frac{3}{8}

Correct answerYour answer

\frac{3}{4}

Correct answerYour answer

\frac{3}{2}

Correct answerYour answer

3

Correct answerYour answer

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AnswerB · $\frac{3}{4}$

$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$ .
$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{3t}{2}\right) \div \frac{dx}{dt} = \frac{3}{2} \div 2t = \frac{3}{4t}$ , which is $\frac{3}{4}$ at $t = 1$ .

Worth noting

$\frac{3}{2}$ differentiates $\frac{dy}{dx}$ with respect to $t$ and stops: it still has to be divided by $\frac{dx}{dt}$ .

Question 79

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For the circle $x^2 + y^2 = 9$ , what is $\frac{d^2y}{dx^2}$ in terms of $y$ ?

- \frac{9}{y^{3}}

Correct answerYour answer

\frac{9}{y^{3}}

Correct answerYour answer

- \frac{1}{y}

Correct answerYour answer

- \frac{9}{y^{2}}

Correct answerYour answer

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AnswerA · $- \frac{9}{y^{3}}$

$2x + 2y\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -\frac{x}{y}$ .
By the quotient rule, $\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}$ .
With $x^2 + y^2 = 9$ , $\frac{d^2y}{dx^2} = -\frac{9}{y^3}$ .

Worth noting

$\frac{9}{y^3}$ loses the minus sign of $-\frac{x}{y}$ .

Question 80

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If $y = x^{2} \log x$ , what is $\frac{d^2y}{dx^2}$ ?

x \left(2 \log x + 1\right)

Correct answerYour answer

- 2 \log x - 3

Correct answerYour answer

x^{2} \left(2 \log x + 1\right)^{2}

Correct answerYour answer

2 \log x + 3

Correct answerYour answer

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AnswerD · $2 \log x + 3$

First derivative: $\frac{dy}{dx} = x \left(2 \log x + 1\right)$ .
Differentiate again: $\frac{d^2y}{dx^2} = 2 \log x + 3$ .

Worth noting

$x \left(2 \log x + 1\right)$ stops at the first derivative; $\frac{d^2y}{dx^2}$ differentiates it once more.

Question 81

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If $x = \cos t$ and $y = \sin t$ , what is $\frac{d^2y}{dx^2}$ ?

\operatorname{cosec}^{2} t

Correct answerYour answer

- \cot t

Correct answerYour answer

\operatorname{cosec}^{3} t

Correct answerYour answer

- \operatorname{cosec}^{3} t

Correct answerYour answer

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AnswerD · $- \operatorname{cosec}^{3} t$

$\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t$ .
$\frac{d}{dt}(-\cot t) = \operatorname{cosec}^2 t$ , and dividing by $\frac{dx}{dt} = -\sin t$ gives $\frac{d^2y}{dx^2} = -\operatorname{cosec}^3 t$ .

Worth noting

$\operatorname{cosec}^2 t$ is the derivative with respect to $t$ , not yet divided by $\frac{dx}{dt}$ .

Question 82

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If $y = e^{-x}\cos x$ , what is $\frac{d^2y}{dx^2}$ ?

- 2 e^{- x} \sin x

Correct answerYour answer

2 e^{- x} \cos x

Correct answerYour answer

e^{- x} \cos x

Correct answerYour answer

2 e^{- x} \sin x

Correct answerYour answer

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AnswerD · $2 e^{- x} \sin x$

$\frac{dy}{dx} = -e^{-x}\cos x - e^{-x}\sin x = -e^{-x}(\cos x + \sin x)$ .
$\frac{d^2y}{dx^2} = e^{-x}(\cos x + \sin x) - e^{-x}(\cos x - \sin x) = 2e^{-x}\sin x$ .

Worth noting

$-2e^{-x}\sin x$ slips a sign: the $-$ of $e^{-x}$ meets the $-$ already in front of the first derivative.

Equations with second derivatives

14 questions

Question 83

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If $y = 3 \sin x + 4 \cos x$ , then $\frac{d^2y}{dx^2} + ky = 0$ . What is $k$ ?

-1

Correct answerYour answer

0

Correct answerYour answer

1

Correct answerYour answer

12

Correct answerYour answer

Show the explanationHide the explanation

AnswerC · $1$

Differentiating twice, each term picks up the factor $-1$ : $\frac{d^2y}{dx^2} = - 3 \sin x - 4 \cos x$ .
That is $-1y$ , so $\frac{d^2y}{dx^2} + 1y = 0$ and $k = 1$ .

Worth noting

$-1$ is what $\frac{d^2y}{dx^2}$ equals as a multiple of $y$ ; moving it across the equals sign changes its sign.

Question 84

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If $y = x^{3}$ and $x^{2}\frac{d^2y}{dx^2} = ky$ , what is $k$ ?

2

Correct answerYour answer

3

Correct answerYour answer

6

Correct answerYour answer

9

Correct answerYour answer

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AnswerC · $6$

$\frac{dy}{dx} = 3x^{2}$ and $\frac{d^2y}{dx^2} = 6x$ .
$x^{2} \cdot 6x = 6x^{3} = 6y$ , so $k = 6$ .

Worth noting

3 comes from the first derivative: $x\frac{dy}{dx} = 3y$ .

Question 85

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If $y = e^{2 x} + e^{- 2 x}$ and $\frac{d^2y}{dx^2} = ky$ , what is $k$ ?

-4

Correct answerYour answer

2

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

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AnswerC · $4$

$\frac{dy}{dx} = 2 e^{2 x} - 2 e^{- 2 x}$ .
$\frac{d^2y}{dx^2} = 4 \left(e^{4 x} + 1\right) e^{- 2 x}$ , which is $4y$ .

Worth noting

2 comes from differentiating once; the factor 2 comes out a second time.

Question 86

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Which relation does $y = xe^x$ satisfy?

\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0

Correct answerYour answer

\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0

Correct answerYour answer

\frac{d^2y}{dx^2} - y = 0

Correct answerYour answer

\frac{d^2y}{dx^2} - 2\frac{dy}{dx} = 0

Correct answerYour answer

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AnswerB · $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$

$\frac{dy}{dx} = (x + 1)e^x$ and $\frac{d^2y}{dx^2} = (x + 2)e^x$ .
$(x + 2) - 2(x + 1) + x = 0$ , so $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$ . The others leave $4x + 4$ , $2$ and $-x$ times $e^x$ .

Question 87

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Which of the following statements are true?

I. $y = e^{3x}$ satisfies $\frac{d^2y}{dx^2} = 9y$ .

II. $y = \sin 3x$ satisfies $\frac{d^2y}{dx^2} = 9y$ .

III. $y = \cos 3x$ satisfies $\frac{d^2y}{dx^2} + 9y = 0$ .

IV. $y = e^{-3x}$ satisfies $\frac{d^2y}{dx^2} = -9y$ .

I, II and III onlyCorrect answerYour answerI and III onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answerII and IV onlyCorrect answerYour answer

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AnswerB · I and III only

Exponentials keep their sign: $e^{3x}$ and $e^{-3x}$ both give $\frac{d^2y}{dx^2} = 9y$ , so I is true and IV is false.
Sine and cosine change sign: both give $\frac{d^2y}{dx^2} = -9y$ , so II is false and III is true.

Question 88

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Match each function in List I with its second derivative in List II.

List I	List II
A. $y = e^{2x}$	I. $2$
B. $y = \sin 2x$	II. $4y$
C. $y = x^2$	III. $-\frac{1}{x^2}$
D. $y = \log x$	IV. $-4y$

A-IV, B-II, C-I, D-IIICorrect answerYour answerA-II, B-IV, C-III, D-ICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answerA-II, B-IV, C-I, D-IIICorrect answerYour answer

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AnswerD · A-II, B-IV, C-I, D-III

$e^{2x}$ gives $4e^{2x} = 4y$ , and $\sin 2x$ gives $-4\sin 2x = -4y$ .
$x^2$ gives 2, and $\log x$ gives $\frac{1}{x}$ , then $-\frac{1}{x^2}$ .

Question 89

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If $y = 2 \sin 2x - 5 \cos 2x$ , then $\frac{d^2y}{dx^2} + ky = 0$ . What is $k$ ?

-4

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4

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5

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8

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AnswerB · $4$

Differentiating twice, each term picks up the factor $-4$ : $\frac{d^2y}{dx^2} = - 8 \sin 2x + 20 \cos 2x$ .
That is $-4y$ , so $\frac{d^2y}{dx^2} + 4y = 0$ and $k = 4$ .

Worth noting

$-4$ is what $\frac{d^2y}{dx^2}$ equals as a multiple of $y$ ; moving it across the equals sign changes its sign.

Question 90

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If $y = e^{ax}$ and $\frac{d^2y}{dx^2} = 16y$ , what are the possible values of $a$ ?

4

onlyCorrect answerYour answer

16

onlyCorrect answerYour answer

4

-4

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2

-2

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AnswerC · $4$ or $-4$

$\frac{d^2y}{dx^2} = a^2e^{ax} = a^2y$ .
So $a^2 = 16$ , and $a = 4$ or $a = -4$ .

Worth noting

$e^{-4x}$ works too: $(-4)^2 = 16$ .

Question 91

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If $y = \log x$ , what is $x\frac{d^2y}{dx^2} + \frac{dy}{dx}$ ?

0

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\frac{2}{x}

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- \frac{2}{x}

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\frac{1}{x}

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AnswerA · $0$

$\frac{dy}{dx} = \frac{1}{x}$ and $\frac{d^2y}{dx^2} = -\frac{1}{x^{2}}$ .
$x \cdot \left(-\frac{1}{x^{2}}\right) + \frac{1}{x} = 0$ .

Worth noting

$\frac{2}{x}$ takes $\frac{d^2y}{dx^2} = +\frac{1}{x^{2}}$ , losing the minus sign of $\frac{d}{dx}x^{-1}$ .

Question 92

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If $y = Ae^{2x} + Be^{3x}$ , what is $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$ ?

0

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y

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6y

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5\frac{dy}{dx}

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AnswerA · $0$

Each term $e^{mx}$ gives $(m^2 - 5m + 6)e^{mx}$ .
$m^2 - 5m + 6 = (m - 2)(m - 3)$ is 0 for $m = 2$ and $m = 3$ , so the expression is 0 for every $A$ and $B$ .

Question 93

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If $e^y(x + 1) = 1$ , which of these is true?

\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^2

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\frac{d^2y}{dx^2} = \frac{dy}{dx}

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\frac{d^2y}{dx^2} = e^y

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\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2

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AnswerD · $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$

$e^y = \frac{1}{x + 1}$ , so $y = -\log(x + 1)$ and $\frac{dy}{dx} = -\frac{1}{x + 1}$ .
$\frac{d^2y}{dx^2} = \frac{1}{(x + 1)^2}$ , which is $\left(-\frac{1}{x + 1}\right)^2 = \left(\frac{dy}{dx}\right)^2$ .

Worth noting

The minus sign belongs to the first derivative, $-\frac{1}{x + 1}$ , and squaring removes it.

Question 94

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If $y = 2\cos(\log x) - 5\sin(\log x)$ , what is $x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ ?

-y

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y

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0

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2y

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AnswerA · $-y$

$x\frac{dy}{dx} = -2\sin(\log x) - 5\cos(\log x)$ .
Differentiating that: $x\frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{-2\cos(\log x) + 5\sin(\log x)}{x} = -\frac{y}{x}$ .
Multiplying by $x$ : $x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} = -y$ .

Worth noting

0 is what you get only after adding $y$ : $x^2y'' + xy' + y = 0$ .

Question 95

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$y = e^{x} + e^{4x}$ satisfies $\frac{d^2y}{dx^2} + p\frac{dy}{dx} + qy = 0$ . What are $p$ and $q$ ?

p = 5,\ q = 4

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p = -5,\ q = 4

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p = -5,\ q = -4

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p = -4,\ q = 5

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AnswerB · $p = -5,\ q = 4$

$\frac{dy}{dx} = e^{x} + 4e^{4x}$ and $\frac{d^2y}{dx^2} = e^{x} + 16e^{4x}$ .
Each exponential must vanish on its own: $1 + p + q = 0$ and $16 + 4p + q = 0$ . Subtracting, $15 + 3p = 0$ , so $p = -5$ and $q = 4$ .

Worth noting

$p = 5$ flips the sign: the powers 1 and 4 are the roots of $m^{2} + pm + q = 0$ , so $p = -(1 + 4)$ and $q = 1 \times 4$ .

Question 96

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$y = e^{x}(A\cos x + B\sin x)$ , with $A$ and $B$ constants. Which relation does $y$ satisfy?

\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0

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\frac{d^2y}{dx^2} - 2\frac{dy}{dx} - 2y = 0

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\frac{d^2y}{dx^2} + y = 0

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\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

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AnswerD · $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

$\frac{dy}{dx} = y + e^{x}(-A\sin x + B\cos x)$ , so $e^{x}(-A\sin x + B\cos x) = \frac{dy}{dx} - y$ .
Differentiating again: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + \left(\frac{dy}{dx} - y\right) - y$ , that is $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$ .

Worth noting

$\frac{d^2y}{dx^2} + y = 0$ is the relation for $A\cos x + B\sin x$ alone; the factor $e^{x}$ adds the middle term.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.