JEE Mathematics Methods of Differentiation MCQs Set 06

Answer: (d) \( -\frac35 \)

Explanation:
1. Put \( x=3\tan\theta \): \( y=\cos^{-1}(\cos2\theta)=2|\theta| \).
2. For \( x<0 \): \( y=-2\tan^{-1}\frac x3 \).
3. \( y'=\frac{-6}{9+x^2} \); at \( x=-1 \): \( -\frac{6}{10}=-\frac35 \).

Teacher's Note:
1. For negative x the inverse cosine returns the positive angle, so the sign flips.
2. Always check the sign of x.

Answer: (a) \( 4x \)

Explanation:
1. \( x>0 \) forces \( t>0 \), so \( x=t \) and \( y=2t^2=2x^2 \).
2. Hence \( y'(x)=4x \).

Teacher's Note:
1. Remove the modulus by fixing the sign of t.
2. Then eliminate t.

Answer: (b) \( \mathrm{cosec}\frac\pi4 \)

Explanation:
1. \( f(x)=\frac{\sin32x}{32\sin x} \).
2. \( f'(x)=\frac{32\cos32x\sin x\cdot32-32\sin32x\cos x}{(32\sin x)^2} \); at \( x=\frac\pi4 \), \( \sin8\pi=0 \), \( \cos8\pi=1 \).
3. \( f'\left(\frac\pi4\right)=\frac{1}{\sin\frac\pi4}=\mathrm{cosec}\frac\pi4 \).

Teacher's Note:
1. Multiply and divide by sin x to use the double-angle product.
2. Most terms vanish at the point.

Answer: (d) \( \frac{\log(e/2)}{2\sqrt2(e^2-1)} \)

Explanation:
1. Let both sides equal \( R \): \( e^R=x+R \) and \( y^R=R \).
2. At \( x=e^2-2 \), \( R=2 \) and \( y=\sqrt2 \).
3. From \( e^RR'=1+R' \): \( R'=\frac{1}{e^2-1} \).
4. From \( R\log y=\log R \): \( y'=\frac yR R'\left(\frac1R-\log y\right)=\frac{\sqrt2}{2}\cdot\frac{1}{e^2-1}\cdot\frac{1-\log2}{2} \), which equals \( \frac{\log(e/2)}{2\sqrt2(e^2-1)} \).

Teacher's Note:
1. Both infinite expressions have the same value R.
2. Differentiate each defining relation separately.

Answer: (c) \( -\cos2x \)

Explanation:
1. \( y=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}=1-\sin x\cos x \).
2. \( y=1-\frac12\sin2x \), so \( y'=-\cos2x \).

Teacher's Note:
1. Rewrite cot and tan in sines and cosines.
2. The sum of cubes factors and cancels the denominator.

Answer: (d) \( \sin x \)

Explanation:
1. The general solution of \( f''=-f \) is \( A\cos x+B\sin x \).
2. \( f(0)=0 \) gives \( A=0 \); \( f'(0)=1 \) gives \( B=1 \).
3. So \( f(x)=\sin x \).

Teacher's Note:
1. Check candidates directly: only sin x satisfies f'' = -f.
2. Verify the initial conditions.

Answer: (b) \( \frac{b}{a(2y+b)} \)

Explanation:
1. The tail repeats: \( y=\frac{x}{a+\frac{x}{b+y}} \).
2. Clearing fractions: \( ay^2+aby=bx \), so \( x=\frac{ay(y+b)}{b} \).
3. \( \frac{dx}{dy}=\frac{a(2y+b)}{b} \), hence \( \frac{dy}{dx}=\frac{b}{a(2y+b)} \).

Teacher's Note:
1. Spot the repeating pattern of two levels.
2. Express x in terms of y and take the reciprocal.

Answer: (b) \( \left(\frac\pi4\right)^{1/\sqrt2}\left(\frac{\sqrt2}{2}\ln\frac4\pi-\frac{2\sqrt2}{\pi}\right) \)

Explanation:
1. Near \( x=-\frac\pi4 \): \( f=(-x)^{-\sin x} \), so \( \ln f=-\sin x\ln(-x) \).
2. \( \frac{f'}{f}=-\cos x\ln(-x)-\frac{\sin x}{x} \).
3. At \( x=-\frac\pi4 \): \( -\frac{\sqrt2}{2}\ln\frac\pi4-\frac{\sqrt2}{2}\cdot\frac4\pi=\frac{\sqrt2}{2}\ln\frac4\pi-\frac{2\sqrt2}{\pi} \), and \( f\left(-\frac\pi4\right)=\left(\frac\pi4\right)^{1/\sqrt2} \).

Teacher's Note:
1. Remove the moduli by using the signs at the given point.
2. Then use logarithmic differentiation.

Answer: (c) \( \frac{-n^2}{1+n^2} \)

Explanation:
1. The terms are \( \tan^{-1}(x+1)-\tan^{-1}x \), \( \tan^{-1}(x+2)-\tan^{-1}(x+1) \), and so on, because \( 1+(x+k)(x+k+1) \) is the denominator.
2. The sum telescopes to \( Y=\tan^{-1}(x+n)-\tan^{-1}x \).
3. \( Y'(0)=\frac{1}{1+n^2}-1=\frac{-n^2}{1+n^2} \).

Teacher's Note:
1. Write each term as a difference of two inverse tangents.
2. The intermediate terms cancel.

Answer: (d) 1

Explanation:
1. \( f'(x)=6xe^{x^2}=2xf(x) \), so the first two terms cancel.
2. \( f(0)=3 \) and \( f'(0)=0 \), so the rest is \( \frac13\cdot3-0=1 \).

Teacher's Note:
1. Notice f' = 2x f before substituting anything.
2. The remaining constants are easy.

Answer: (b) \( \frac{12}{(3-2x)^2}\sin\left[\log\frac{2x+3}{3-2x}\right] \)

Explanation:
1. Chain rule: \( y'=f'(u)\cdot u' \) with \( u=\frac{2x+3}{3-2x} \).
2. \( u'=\frac{2(3-2x)+2(2x+3)}{(3-2x)^2}=\frac{12}{(3-2x)^2} \).
3. So \( y'=\frac{12}{(3-2x)^2}\sin(\log u) \).

Teacher's Note:
1. Evaluate f' at u, not at x.
2. Do not forget the derivative of the inner fraction.

Answer: (b) \( \frac{1}{2e} \)

Explanation:
1. \( f(x)=\frac{\ln(\ln x)}{2\ln x} \).
2. \( f'(x)=\frac{\frac{1}{x\ln x}\cdot2\ln x-2\ln(\ln x)\cdot\frac1x}{4\ln^2x} \).
3. At \( x=e \): \( \ln e=1 \), \( \ln(\ln e)=0 \), so \( f'(e)=\frac{2/e}{4}=\frac{1}{2e} \).

Teacher's Note:
1. Change of base first: log base x squared of u equals ln u over 2 ln x.
2. At x = e the inner log vanishes, which simplifies the algebra.

Answer: (a) \( -\frac{d^2y/dx^2}{(dy/dx)^3} \)

Explanation:
1. \( \frac{dx}{dy}=\frac{1}{y'} \).
2. \( \frac{d^2x}{dy^2}=\frac{d}{dx}\left(\frac1{y'}\right)\cdot\frac{dx}{dy}=-\frac{y''}{y'^2}\cdot\frac1{y'}=-\frac{y''}{y'^3} \).

Teacher's Note:
1. Differentiate with respect to x and then multiply by dx/dy.
2. This is a formula worth remembering.

Answer: (b) \( n^2y^2 \)

Explanation:
1. \( \frac{dy}{dx}=n\left(x+\sqrt{x^2-a^2}\right)^{n-1}\left(1+\frac{x}{\sqrt{x^2-a^2}}\right)=\frac{ny}{\sqrt{x^2-a^2}} \).
2. Squaring and multiplying by \( x^2-a^2 \) gives \( n^2y^2 \).

Teacher's Note:
1. The factor 1 + x over root equals (root + x) over root, which reproduces y.
2. Square both sides.

Answer: (d) \( \frac{p}{q(1-x)^{\frac pq+1}} \)

Explanation:
1. With \( s=\frac pq \) the series is \( 1+sx+\frac{s(s+1)}{2!}x^2+\cdots=(1-x)^{-s} \).
2. Differentiate: \( s(1-x)^{-s-1}=\frac pq(1-x)^{-\frac pq-1} \).

Teacher's Note:
1. Identify the binomial series for a negative exponent.
2. Then differentiate the closed form.

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

Explanation:
1. Since \( 1+x^2-x>0 \), \( y=\tan^{-1}\frac{1}{1+x(x-1)}=\tan^{-1}x-\tan^{-1}(x-1) \).
2. Differentiate: \( \frac{1}{1+x^2}-\frac{1}{1+(x-1)^2} \).

Teacher's Note:
1. Use the formula tan inverse a minus tan inverse b.
2. Positive argument means cot inverse equals tan inverse of the reciprocal.

Answer: (a) \( \frac{1}{x\sqrt{\sec^{-1}x^2}\sqrt{x^4-1}} \)

Explanation:
1. \( \frac{1}{2\sqrt{\sec^{-1}x^2}}\cdot\frac{1}{x^2\sqrt{x^4-1}}\cdot2x \).
2. Simplify: \( \frac{1}{x\sqrt{\sec^{-1}x^2}\sqrt{x^4-1}} \).

Teacher's Note:
1. Apply the chain rule through three layers.
2. The factor 2x cancels the 2 and one power of x.

Answer: (b) \( \frac14\mathrm{cosech}^4\frac x2 \)

Explanation:
1. Derivative of \( \coth\frac x2 \) is \( -\frac12\mathrm{cosech}^2\frac x2 \).
2. Total: \( -\frac14\mathrm{cosech}^2\frac x2+\frac14\coth^2\frac x2\,\mathrm{cosech}^2\frac x2=\frac14\mathrm{cosech}^2\frac x2\left(\coth^2\frac x2-1\right) \).
3. Since \( \coth^2u-1=\mathrm{cosech}^2u \), the result is \( \frac14\mathrm{cosech}^4\frac x2 \).

Teacher's Note:
1. Use the identity coth^2 - 1 = cosech^2.
2. The answer is a pure fourth power.

Answer: (d) \( (n+1)\mathrm{cosec}^2(n+1)x-\mathrm{cosec}^2x \)

Explanation:
1. Use \( \frac{1}{\sin kx\sin(k+1)x}=\frac{\cot kx-\cot(k+1)x}{\sin x} \).
2. The sum telescopes: \( y=\cot x-\cot(n+1)x \).
3. \( y'=-\mathrm{cosec}^2x+(n+1)\mathrm{cosec}^2(n+1)x \).

Teacher's Note:
1. The identity comes from sin((k+1)x - kx) = sin x.
2. Telescoping removes all middle terms.

Answer: (d) \( \frac12 \)

Explanation:
1. Replace \( x \) by \( \frac1x \): \( 3f(\frac1x)-2f(x)=\frac1x \).
2. Eliminating \( f(1/x) \): \( 5f(x)=3x+\frac2x \), so \( f(x)=\frac{3x+2/x}{5} \).
3. \( f'(x)=\frac{3-2/x^2}{5} \); \( f'(2)=\frac{3-0.5}{5}=\frac12 \).

Teacher's Note:
1. Functional equations with x and 1/x are solved by the substitution trick.
2. Differentiate only after finding f explicitly.

Answer: (a) \( 2x[1+\tan(\log x)]+x\sec^2(\log x) \)

Explanation:
1. \( \log x=\tan^{-1}\frac{y-x^2}{x^2} \), so \( y=x^2+x^2\tan(\log x) \).
2. \( y'=2x+2x\tan(\log x)+x^2\sec^2(\log x)\cdot\frac1x \).
3. \( =2x[1+\tan(\log x)]+x\sec^2(\log x) \).

Teacher's Note:
1. Solve for y explicitly.
2. The chain factor 1/x reduces x^2 to x in the last term.

Answer: (b) 2

Explanation:
1. \( f(-1)=-1-2-3+4=-2 \), so \( f^{-1}(-2)=-1 \).
2. \( f'(x)=5x^4+6x^2+3 \), \( f'(-1)=14 \).
3. \( \frac{d}{dx}f^{-1}(-2)=\frac1{14} \), and \( 28\times\frac1{14}=2 \).

Teacher's Note:
1. Find the point on f that maps to the given value.
2. Use the reciprocal rule for inverse functions.

Answer: (c) 1

Explanation:
1. \( f(x)=\frac{\sin2^nx}{2^n\sin x} \).
2. \( f'(x)=\frac{2^n\cos2^nx}{2^n\sin x}-\frac{\sin2^nx\cos x}{2^n\sin^2x} \).
3. At \( x=\frac\pi2 \): \( \cos(2^{n-1}\pi)=1 \), \( \sin(2^{n-1}\pi)=0 \), \( \sin\frac\pi2=1 \), so \( f'=1 \).

Teacher's Note:
1. Multiply and divide by sin x to use the double angle formula repeatedly.
2. For n > 1, 2^(n-1) is an integer so the sine term vanishes.

Answer: (d) \( \mathrm{cosec}(g(x)) \)

Explanation:
1. \( g'(x)=\frac{1}{f'(g(x))}=\frac{1}{\sin g(x)} \).
2. This equals \( \mathrm{cosec}(g(x)) \).

Teacher's Note:
1. Inverse-function derivative: reciprocal of f' at g(x).
2. No minus sign appears.