If x = t^2 and y = t^3, what is d^2y/dx^2 at t = 1?

Derivatives & Second-Order Derivatives · Question 78 of 96

Question 78

HardOriginal question

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

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

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

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.