Chain Rule, Log Differentiation and Rate of Change for CUET Maths

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

1. Outside $\sin u$, inside $u = x^{2}$: $\cos u \cdot u'$.

2. $\cos(x^{2}) \cdot 2x$.

AnswerC · $10$

1. $f'(x) = 5(2x + 1)^{4} \cdot 2 = 10(2x + 1)^{4}$.

2. $f'(0) = 10$.

AnswerD · $24 x \left(3 x^{2} - 5\right)^{3}$

1. $\frac{d}{dx} u^{4} = 4u^{3} u'$ with $u = 3x^{2} - 5$, $u' = 6x$.

2. $4(3x^{2} - 5)^{3} \cdot 6x = 24x(3x^{2} - 5)^{3}$.

AnswerB · 4 m

1. At $t = 0$: $s = \sqrt{16}$.

2. 4 m.

AnswerB · 0.6 m/s

1. $\frac{3}{\sqrt{9 + 16}} = \frac{3}{5}$.

2. 0.6 m/s.

AnswerD · $24$

1. $f'(x) = 3(x^{2} + 1)^{2} \cdot 2x$.

2. At 1: $3 \times 4 \times 2 = 24$.

AnswerD · $3 \tan 3x \sec 3x$

1. $\frac{d}{dx}\sec u = \sec u\tan u \cdot u'$ with $u = 3x$.

2. $3\sec 3x\tan 3x$.

AnswerB · $-12$

1. $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$.

2. At 1: $f'(g(1)) \cdot g'(1) = f'(2) \times (-4) = 3 \times (-4) = -12$.

AnswerA · $-\frac{12x}{(3x^{2} + 1)^{3}}$

1. Write it as $u^{-2}$ with $u = 3x^{2} + 1$: the derivative is $-2u^{-3} \cdot u'$, $u' = 6x$.

2. $-2(3x^{2} + 1)^{-3} \cdot 6x = -\frac{12x}{(3x^{2} + 1)^{3}}$.

AnswerC · $-4\operatorname{cosec}^{2} 4x$

1. $\frac{d}{dx}\cot u = -\operatorname{cosec}^{2} u \cdot u'$ with $u = 4x$.

2. $-\operatorname{cosec}^{2} 4x \cdot 4 = -4\operatorname{cosec}^{2} 4x$.

AnswerA · Only I

1. I: $2\sin x\cos x = \sin 2x$. True.

2. II: the inner $x^{2}$ gives a factor $2x$, so it is $2x\cos(x^{2})$. False.

AnswerA · $\frac{3 x^{2}}{2 \sqrt{x^{3} + 1}}$

1. $\frac{d}{dx}\sqrt{u} = \frac{u'}{2\sqrt{u}}$ with $u = 1 + x^{3}$.

2. $\frac{3x^{2}}{2\sqrt{1 + x^{3}}}$.

AnswerC · $- 6 \sin 2x \cos^{2} 2x$

1. Three layers: $u^{3}$, then $\cos v$, then $v = 2x$.

2. $3\cos^{2} 2x \cdot (-\sin 2x) \cdot 2 = -6\cos^{2} 2x\,\sin 2x$.

AnswerB · $- \sin x \cos{\left(\cos x \right)}$

1. Outside $\sin u$, inside $u = \cos x$, $u' = -\sin x$.

2. $\cos(\cos x) \cdot (-\sin x)$.

AnswerA · At $t = 0$

1. At rest means $\frac{t}{\sqrt{t^{2} + 16}} = 0$.

2. That happens only at $t = 0$.

AnswerB · $\frac{4}{5}$

1. $f'(x) = \frac{2x}{2\sqrt{x^{2} + 9}} = \frac{x}{\sqrt{x^{2} + 9}}$.

2. At 4: $\frac{4}{5}$.

AnswerC · $2$

1. $f'(x) = 2\sin 2x \cdot \cos 2x \cdot 2 = 2\sin 4x$.

2. At $\frac{\pi}{8}$: $2\sin \frac{\pi}{2} = 2$.

AnswerA · $\frac{\cos x}{2 \sqrt{\sin x}}$

1. $\frac{d}{dx}\sqrt{u} = \frac{u'}{2\sqrt{u}}$ with $u = \sin x$.

2. $\frac{\cos x}{2\sqrt{\sin x}}$.

AnswerB · $3 \sin^{2} x \cos x$

1. Outside $u^{3}$, inside $u = \sin x$.

2. $3\sin^{2} x \cdot \cos x$.

AnswerD · $- \frac{\sin{\left(\sqrt{x} \right)}}{2 \sqrt{x}}$

1. $-\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$.

2. $-\frac{\sin\sqrt{x}}{2\sqrt{x}}$.

AnswerA · $\frac{t}{\sqrt{t^{2} + 16}}$

1. $\frac{d}{dt}\sqrt{u} = \frac{u'}{2\sqrt{u}}$ with $u = t^{2} + 16$.

2. $\frac{2t}{2\sqrt{t^{2} + 16}} = \frac{t}{\sqrt{t^{2} + 16}}$.

AnswerD · $\frac{\pi}{180}\cos x^{\circ}$

1. $x^{\circ} = \frac{\pi x}{180}$ radians, so $\sin x^{\circ} = \sin\frac{\pi x}{180}$.

2. Chain rule: $\cos\frac{\pi x}{180} \cdot \frac{\pi}{180} = \frac{\pi}{180}\cos x^{\circ}$.

AnswerB · Both A and R are true, but R is not the correct explanation of A.

1. A is true: $f'(x) = 2x\cos(x^{2})$, and $f'(-x) = -2x\cos(x^{2}) = -f'(x)$.

2. R is true (differentiate $g(-x) = -g(x)$ by the chain rule), but $\sin(x^{2})$ is even, so A comes from the rule for even functions, not from R.

AnswerA · $\frac{3}{4}$

1. $F'(x) = f'(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$.

2. At 16: $\sqrt{16} = 4$, so $F'(16) = f'(4) \cdot \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$.

AnswerA · $\frac{1}{24}$

1. $f'(x) = \frac{1}{2\sqrt{1 + \sqrt{x}}} \cdot \frac{1}{2\sqrt{x}}$, one factor per layer.

2. At 9: $\sqrt{x} = 3$ and $\sqrt{1 + 3} = 2$, so $\frac{1}{4} \cdot \frac{1}{6} = \frac{1}{24}$.

AnswerC · A-III, B-I, C-IV, D-II

1. A: $4(1 - 3x)^{3}(-3)$ at 0 is $-12$. B: $\frac{8}{2\sqrt{1 + 8x}}$ at 0 is 4.

2. C: $3\sec^{2} 3x$ at 0 is 3. D: $-3(1 + 2x)^{-4} \cdot 2$ at 0 is $-6$.