JEE Mathematics Methods of Differentiation MCQs Set 02

Answer: (d) (16, 15)

Explanation:
1. Multiply and divide by \( (x-1) \): the product equals \( \frac{x^{16}-1}{x-1} \).
2. Differentiate by the quotient rule: \( f'(x)=\frac{16x^{15}(x-1)-(x^{16}-1)}{(x-1)^2}=\frac{15x^{16}-16x^{15}+1}{(x-1)^2} \).
3. Hence \( p=16 \), \( q=15 \).

Teacher's Note:
1. Spotting that the product telescopes saves a long product-rule expansion.
2. Always compare the derived numerator with the given form term by term.

Answer: (d) \( \frac{\sin(\log x)}{2x} \)

Explanation:
1. Chain rule: \( 2\sin(\log\sqrt x)\cos(\log\sqrt x)\cdot\frac{1}{\sqrt x}\cdot\frac{1}{2\sqrt x} \).
2. Use \( 2\sin\theta\cos\theta=\sin2\theta \) with \( 2\log\sqrt x=\log x \).
3. Result: \( \frac{\sin(\log x)}{2x} \).

Teacher's Note:
1. Simplify the inside first: \( \log\sqrt x=\tfrac12\log x \).
2. A double-angle identity often collapses the answer into one term.

Answer: (a) 0

Explanation:
1. \( [x-\pi] \) is an integer, so \( [x-\pi]\pi \) is an integral multiple of \( \pi \).
2. Therefore \( \tan([x-\pi]\pi)=0 \) and \( f(x)=0 \) for every real \( x \).
3. A constant function has derivative \( 0 \).

Teacher's Note:
1. Check whether a messy-looking expression is actually constant before differentiating.
2. Here the tangent of an integral multiple of pi kills the whole function.

Answer: (c) 2

Explanation:
1. Expand along the first row: \( f(x)=2\cos x(4\cos^2x-1)-\left(x-\frac\pi2\right)2\cos x=\cos x\,g(x) \) with \( g(x)=8\cos^2x-2x-2+\pi \).
2. \( f'(x)=-\sin x\,g(x)+\cos x\,g'(x) \) and \( g'(x)=-16\sin x\cos x-2 \).
3. At \( x=\pi \): \( \sin\pi=0 \), \( \cos\pi=-1 \), \( g'(\pi)=-2 \), so \( f'(\pi)=(-1)(-2)=2 \).

Teacher's Note:
1. A determinant of functions can be expanded first and then differentiated like any other function.
2. Substitute the point only after differentiating.

Answer: (c) \( -\tan\theta \)

Explanation:
1. \( \frac{dx}{d\theta}=-3a\cos^2\theta\sin\theta \) and \( \frac{dy}{d\theta}=3a\sin^2\theta\cos\theta \).
2. \( \frac{dy}{dx}=\frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\tan\theta \).

Teacher's Note:
1. For a parametric curve, divide the two derivatives with respect to the parameter.
2. Cancel common factors carefully before naming the answer.

Answer: (b) \( -\frac{ax+hy+g}{hx+by+f} \)

Explanation:
1. Differentiate implicitly: \( 2ax+2hy+2hx\,y'+2by\,y'+2g+2f\,y'=0 \).
2. Collect \( y' \): \( y'(hx+by+f)=-(ax+hy+g) \).
3. So \( y'=-\frac{ax+hy+g}{hx+by+f} \).

Teacher's Note:
1. This is the general conic result and is worth remembering.
2. The partial-derivative form \( -F_x/F_y \) gives it at once.

Answer: (c) \( \frac{1}{\sqrt{1-x^2}} \)

Explanation:
1. \( h(x)=e^{\sin^{-1}x} \).
2. \( h'(x)=e^{\sin^{-1}x}\cdot\frac{1}{\sqrt{1-x^2}} \).
3. Dividing by \( h(x) \) gives \( \frac{1}{\sqrt{1-x^2}} \).

Teacher's Note:
1. The ratio h'/h is the logarithmic derivative of h.
2. For an exponential, it is just the derivative of the exponent.

Answer: (b) 6

Explanation:
1. Differentiate \( f(x+h)=f(x)f(h) \) with respect to \( h \) at \( h=0 \): \( f'(x)=f(x)f'(0) \).
2. At \( x=5 \): \( f'(5)=2\times3=6 \).

Teacher's Note:
1. The relation f(x+y)=f(x)f(y) describes an exponential function.
2. Then f'(x)=f'(0)f(x) at every point.

Answer: (d) \( -\frac{1}{2(1+x^2)} \)

Explanation:
1. Put \( x=\tan\theta \): \( \frac{\tan\theta}{\sec\theta-1}=\frac{\sin\theta}{1-\cos\theta}=\cot\frac\theta2 \).
2. \( y=\tan^{-1}\left(\cot\frac\theta2\right)=\frac\pi2-\frac\theta2=\frac\pi2-\frac12\tan^{-1}x \).
3. \( \frac{dy}{dx}=-\frac{1}{2(1+x^2)} \).

Teacher's Note:
1. A root of the form \( \sqrt{1+x^2} \) suggests \( x=\tan\theta \).
2. Reduce to the half-angle cotangent, then use \( \tan^{-1}(\cot\alpha)=\frac\pi2-\alpha \).

Answer: (d) \( \frac yx \)

Explanation:
1. Write \( \frac yx=\sin^{-1}\left(\frac xy\right) \), a relation involving only \( \frac yx \).
2. Hence \( \frac yx \) is constant, i.e. \( y=cx \).
3. So \( \frac{dy}{dx}=c=\frac yx \).

Teacher's Note:
1. If an equation can be written purely in terms of y/x, then y/x is constant.
2. A homogeneous relation of this type always gives dy/dx = y/x.

Answer: (b) 2

Explanation:
1. Every term has total degree \( 22 \), so the equation is homogeneous of degree \( 22 \).
2. Dividing by \( x^{22} \) gives a relation in \( u=\frac yx \) only, so \( \frac yx \) is constant along the curve.
3. Hence \( \frac{dy}{dx}=\frac yx=\frac21=2 \).

Teacher's Note:
1. Counting the total degree of every term identifies a homogeneous equation.
2. Homogeneous curves of this kind behave like straight lines through the origin.

Answer: (c) \( -\frac yx \)

Explanation:
1. \( f(xy)+g(xy) \) is a function of the product \( xy \), so \( xy \) is constant.
2. Differentiate \( xy=c \): \( y+xy'=0 \), giving \( y'=-\frac yx \).

Teacher's Note:
1. Whenever an equation involves only the product xy, treat xy as a constant.
2. The same idea gives y/x for a function of y/x.

Answer: (c) \( \frac{\cos\sqrt x}{4\sqrt{x\sin\sqrt x}} \)

Explanation:
1. Chain rule through three layers: \( \frac{1}{2\sqrt{\sin\sqrt x}}\cdot\cos\sqrt x\cdot\frac{1}{2\sqrt x} \).
2. Combine the roots: \( \frac{\cos\sqrt x}{4\sqrt{x\sin\sqrt x}} \).

Teacher's Note:
1. Differentiate layer by layer from the outside in.
2. Multiply the square roots together at the end.

Answer: (d) \( y\,a^x(\log a)^2 \)

Explanation:
1. Take logs twice: \( \log y=a^x\log a \).
2. Differentiate: \( \frac{y'}{y}=a^x(\log a)\cdot\log a \).
3. So \( y'=y\,a^x(\log a)^2 \).

Teacher's Note:
1. A tower of exponentials needs the logarithm applied once per level.
2. Each level contributes one factor of log a.

Answer: (d) \( \frac{1}{x\log x\,\log(\log x)} \)

Explanation:
1. \( \frac{1}{\log(\log x)}\cdot\frac{1}{\log x}\cdot\frac1x \).

Teacher's Note:
1. Peel the logarithms off one at a time, outermost first.
2. Each layer contributes one reciprocal factor.

Answer: (b) 1

Explanation:
1. \( \cot\left(\frac\pi2-x\right)=\tan x \).
2. So \( y=\tan^{-1}(\tan x)=x \) (principal range), and \( y'=1 \).

Teacher's Note:
1. Reduce the co-function first.
2. Remember that \( \tan^{-1}(\tan x)=x \) only on the principal interval.

Answer: (a) \( \frac12 \)

Explanation:
1. \( \sec x+\tan x=\frac{1+\sin x}{\cos x}=\tan\left(\frac\pi4+\frac x2\right) \).
2. So \( y=\frac\pi4+\frac x2 \) and \( y'=\frac12 \).

Teacher's Note:
1. The expression sec x + tan x equals tan(pi/4 + x/2).
2. Recognising this identity turns the problem into a linear function.

Answer: (c) \( \frac{3}{2(1+x)\sqrt x} \)

Explanation:
1. Put \( \sqrt x=\tan\theta \): the fraction is \( \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\tan3\theta \).
2. So \( y=3\theta=3\tan^{-1}\sqrt x \).
3. \( y'=\frac{3}{1+x}\cdot\frac{1}{2\sqrt x} \).

Teacher's Note:
1. The expression (3t - t^3)/(1 - 3t^2) is tan 3A in disguise.
2. Substitute sqrt x = tan theta to expose it.

Answer: (d) \( y(\log\sin x+x\cot x) \)

Explanation:
1. \( \log y=x\log\sin x \).
2. \( \frac{y'}{y}=\log\sin x+x\cdot\frac{\cos x}{\sin x} \).
3. \( y'=y(\log\sin x+x\cot x) \).

Teacher's Note:
1. The product rule applies to x times log(sin x).
2. The derivative of log(sin x) is cot x.

Answer: (d) \( \frac{\log x}{(1+\log x)^2} \)

Explanation:
1. Take logs: \( y\log x=x-y \), so \( y=\frac{x}{1+\log x} \).
2. \( \frac{dy}{dx}=\frac{(1+\log x)-x\cdot\frac1x}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2} \).

Teacher's Note:
1. Solving for y explicitly first avoids implicit differentiation.
2. Use the quotient rule carefully; the constant terms cancel.

Answer: (a) \( \tanh x \)

Explanation:
1. \( \frac{1}{\cosh x}\cdot\sinh x=\tanh x \).

Teacher's Note:
1. The derivative of cosh x is sinh x.
2. The ratio sinh/cosh is tanh.

Answer: (c) \( \frac{3}{\sqrt{1+9x^2}} \)

Explanation:
1. \( \frac{d}{du}\sinh^{-1}u=\frac{1}{\sqrt{1+u^2}} \).
2. With \( u=3x \) the chain factor is \( 3 \), giving \( \frac{3}{\sqrt{1+9x^2}} \).

Teacher's Note:
1. Do not skip the chain-rule factor 3.
2. The formula for sinh inverse is the same as the circular one with a plus sign.

Answer: (a) \( \frac t2 \)

Explanation:
1. \( \frac{dx}{dt}=a(1+\cos t) \) and \( \frac{dy}{dt}=a\sin t \).
2. \( \frac{dx}{dy}=\frac{1+\cos t}{\sin t}=\frac{2\cos^2\frac t2}{2\sin\frac t2\cos\frac t2}=\cot\frac t2 \).
3. So \( p=\frac t2 \).

Teacher's Note:
1. Half-angle identities simplify cycloid derivatives.
2. Compare the result with the given form cot p directly.

Answer: (c) \( \tan t \)

Explanation:
1. \( \frac{dx}{dt}=a\left(-\sin t+\frac{1}{\sin t}\right)=\frac{a\cos^2t}{\sin t} \), since \( \frac{d}{dt}\log\tan\frac t2=\frac{1}{\sin t} \).
2. \( \frac{dy}{dt}=a\cos t \).
3. \( \frac{dy}{dx}=\frac{a\cos t\sin t}{a\cos^2t}=\tan t \).

Teacher's Note:
1. This is the parametric form of the tractrix.
2. The derivative of log tan(t/2) is cosec t.

Answer: (b) \( a^x\log_ea\sqrt{1-x^2} \)

Explanation:
1. \( \frac{d(a^x)}{d(\sin^{-1}x)}=\frac{a^x\log a}{1/\sqrt{1-x^2}}=a^x\log_ea\sqrt{1-x^2} \).

Teacher's Note:
1. The derivative of u with respect to v is (du/dx)/(dv/dx).
2. The reciprocal of 1/sqrt(1-x^2) moves the root to the numerator.