JEE Mathematics Methods of Differentiation MCQs Set 04

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

Explanation:
1. Put \( x=\cos2\theta \): the fraction is \( \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\tan\left(\frac\pi4-\theta\right) \).
2. The function is \( \frac\pi2-\left(\frac\pi4-\theta\right)=\frac\pi4+\frac12\cos^{-1}x \).
3. Derivative: \( -\frac{1}{2\sqrt{1-x^2}} \).

Teacher's Note:
1. With square roots of 1 plus x and 1 minus x, use x = cos 2 theta.
2. Remember the final derivative of cos inverse.

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

Explanation:
1. Put \( x=\cos2\theta \) with \( 0<\theta<\frac\pi4 \): the bracket is \( \frac{\cos\theta+\sin\theta}{\sqrt2}=\cos\left(\frac\pi4-\theta\right) \).
2. So the function is \( \frac\pi2-\left(\frac\pi4-\theta\right)=\frac\pi4+\frac12\cos^{-1}x \).
3. Derivative: \( -\frac{1}{2\sqrt{1-x^2}} \).

Teacher's Note:
1. The restriction 0 < x < 1 fixes the principal value branch.
2. The result agrees with the previous question.

Answer: (c) \( \frac{4\log3}{(3^x+3^{-x})^2} \)

Explanation:
1. Quotient rule with numerator derivative \( \log3(3^x+3^{-x}) \) and denominator derivative \( \log3(3^x-3^{-x}) \).
2. Numerator: \( \log3[(3^x+3^{-x})^2-(3^x-3^{-x})^2]=4\log3 \).
3. Result: \( \frac{4\log3}{(3^x+3^{-x})^2} \).

Teacher's Note:
1. The difference of squares of the sum and difference equals 4.
2. This is the tanh derivative in exponential form.

Answer: (b) \( \mathrm{sech}\,x \)

Explanation:
1. \( \frac{\cosh x}{1+\sinh^2x}=\frac{\cosh x}{\cosh^2x}=\frac{1}{\cosh x} \).

Teacher's Note:
1. Use 1 + sinh^2 x = cosh^2 x.
2. The result is the hyperbolic secant.

Answer: (a) \( x^{\tan x}\left\{\frac{\tan x}{x}+\sec^2x\log x\right\} \)

Explanation:
1. \( \log y=\tan x\log x \).
2. \( \frac{y'}{y}=\sec^2x\log x+\frac{\tan x}{x} \).

Teacher's Note:
1. Product rule on tan x times log x.
2. Multiply back by y.

Answer: (b) \( -(n-1)! \)

Explanation:
1. At \( x=1 \) the first factor vanishes, so only the term where it is differentiated survives: \( (-1)(2-1)(3-1)\cdots(n-1) \).
2. This equals \( -(n-1)! \).

Teacher's Note:
1. In a product, a factor that vanishes at the point kills every term except the one where it is differentiated.
2. The remaining factors multiply to (n-1)!.

Answer: (b) \( -\frac{2}{\sqrt3} \)

Explanation:
1. \( t=\sin x \), so \( y=\log\cos^2x=2\log\cos x \).
2. \( \frac{dy}{dx}=-2\tan x \).
3. At \( t=\frac12 \), \( x=\frac\pi6 \) and \( \tan x=\frac{1}{\sqrt3} \), giving \( -\frac{2}{\sqrt3} \).

Teacher's Note:
1. Eliminating the parameter can be quicker than parametric differentiation.
2. Convert the value of t back to the angle.

Answer: (b) \( \frac{2t}{1-t^2} \)

Explanation:
1. \( \frac{dx}{dt}=\frac{3a(1-t^2)}{(1+t^2)^2} \) and \( \frac{dy}{dt}=\frac{6at}{(1+t^2)^2} \).
2. Dividing: \( \frac{2t}{1-t^2} \).

Teacher's Note:
1. The common denominator cancels in the ratio.
2. This is the folium of Descartes.

Answer: (d) \( -\frac32 \)

Explanation:
1. \( \frac{dy}{dx}=\frac{\sin2t-\sin t}{\cos t-\cos2t} \).
2. \( \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt} \).
3. Evaluating at \( t=\frac\pi2 \) gives \( -\frac32 \).

Teacher's Note:
1. The second derivative needs the t-derivative of dy/dx divided again by dx/dt.
2. Substitute the angle only after forming the full quotient.

Answer: (d) \( -\frac83 \)

Explanation:
1. \( y=2\log\cos x \), so \( \frac{dy}{dx}=-2\tan x \).
2. \( \frac{d^2y}{dx^2}=-2\sec^2x \).
3. At \( x=\frac\pi6 \): \( -2\cdot\frac43=-\frac83 \).

Teacher's Note:
1. Eliminate the parameter when it is easy.
2. sec^2(pi/6) = 4/3.

Answer: (d) \( \frac14 \)

Explanation:
1. First function: \( \frac12\tan^{-1}x \).
2. Second: put \( x=\sin\theta \), giving \( \tan^{-1}(\tan2\theta)=2\sin^{-1}x \).
3. Ratio of derivatives: \( \frac{\frac12\cdot\frac{1}{1+x^2}}{\frac{2}{\sqrt{1-x^2}}} \), which at \( x=0 \) equals \( \frac14 \).

Teacher's Note:
1. Reduce each function by a suitable substitution.
2. Then divide the two derivatives.

Answer: (d) \( \frac{\cos^2(a+y)}{\sin a} \)

Explanation:
1. \( x=\frac{\cos y}{\cos(a+y)} \).
2. \( \frac{dx}{dy}=\frac{-\sin y\cos(a+y)+\cos y\sin(a+y)}{\cos^2(a+y)}=\frac{\sin a}{\cos^2(a+y)} \).
3. So \( \frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a} \).

Teacher's Note:
1. Differentiate x with respect to y when x is easier to isolate.
2. The numerator simplifies by sin(A - B).

Answer: (d) \( \frac{3a^{3x+1}\log a}{2y-1} \)

Explanation:
1. The radical repeats, so \( y^2=a^{3x+1}+y \).
2. Differentiate: \( 2yy'=3a^{3x+1}\log a+y' \).
3. \( y'=\frac{3a^{3x+1}\log a}{2y-1} \).

Teacher's Note:
1. An infinite radical gives a self-referential equation.
2. Differentiate that equation instead of the radical.

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

Explanation:
1. Move one term across and square: \( x^2(1+y)=y^2(1+x) \), so \( (x-y)(x+y+xy)=0 \).
2. As \( x\ne y \), \( x+y+xy=0 \), i.e. \( y=\frac{-x}{1+x} \).
3. \( \frac{dy}{dx}=\frac{-1}{(1+x)^2} \).

Teacher's Note:
1. Squaring and factoring removes the radicals.
2. The condition x not equal to y discards the factor x - y.

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

Explanation:
1. \( \frac{1+\frac{x}{\sqrt{a^2+x^2}}}{x+\sqrt{a^2+x^2}}=\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}\,(x+\sqrt{a^2+x^2})}=\frac{1}{\sqrt{a^2+x^2}} \).

Teacher's Note:
1. This is the standard form of the sinh inverse derivative.
2. The factor x + root cancels from numerator and denominator.

Answer: (a) \( \frac12\sec^2\frac x2 \)

Explanation:
1. \( \frac{\sec x-1}{\sec x+1}=\frac{1-\cos x}{1+\cos x}=\tan^2\frac x2 \).
2. So \( y=\tan\frac x2 \) and \( y'=\frac12\sec^2\frac x2 \).

Teacher's Note:
1. Multiply numerator and denominator by cos x to convert to half-angle form.
2. The root is positive on the stated interval.

Answer: (c) \( -4 \)

Explanation:
1. \( y=\left(\frac{x+1}{3x+1}\right)^2 \).
2. \( y'=2\cdot\frac{x+1}{3x+1}\cdot\frac{(3x+1)-3(x+1)}{(3x+1)^2}=\frac{-4(x+1)}{(3x+1)^3} \).
3. At \( x=0 \): \( -4 \).

Teacher's Note:
1. Use a^(log_a u) = u to remove the exponential.
2. Differentiate the resulting rational function.

Answer: (b) \( \frac{f'(x)}{g'(y)}\cdot\frac{1}{\sqrt{1+(f(x))^2}} \)

Explanation:
1. The relation says \( \sinh g(y)=f(x) \).
2. Differentiate: \( \cosh g(y)\,g'(y)\,y'=f'(x) \).
3. With \( \cosh g(y)=\sqrt{1+f(x)^2} \), \( y'=\frac{f'(x)}{g'(y)\sqrt{1+f(x)^2}} \).

Teacher's Note:
1. Recognise (e^u - e^-u)/2 as sinh u.
2. Use cosh^2 = 1 + sinh^2 to eliminate g(y).

Answer: (c) 0

Explanation:
1. \( \sin^{-1}2x+\cos^{-1}2x=\frac\pi2 \) and \( \tan^{-1}3x+\cot^{-1}3x=\frac\pi2 \).
2. So \( y=\pi \), a constant, and \( y'=0 \).

Teacher's Note:
1. Complementary inverse functions add to pi/2.
2. The sum is therefore constant.

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

Explanation:
1. With \( x=\tan\theta \), \( \frac{\sec\theta-1}{\tan\theta}=\tan\frac\theta2 \).
2. \( y=\frac12\tan^{-1}x \), so \( y'=\frac{1}{2(1+x^2)} \).

Teacher's Note:
1. Half-angle tangent again.
2. Do not forget the factor one half.

Answer: (b) \( -\frac12 \)

Explanation:
1. Put \( x=\cos2\theta \): \( \sqrt{\frac{1+x}{1-x}}=\cot\theta \), so \( \cot^{-1}(\cot\theta)=\theta \).
2. \( \sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-x}{2} \).
3. Derivative: \( -\frac12 \).

Teacher's Note:
1. Use x = cos 2 theta for (1 + x)/(1 - x).
2. Then apply a half-angle identity.

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

Explanation:
1. Put \( x=\sin\theta \), \( \cos\alpha=\frac5{13} \), \( \sin\alpha=\frac{12}{13} \): the bracket is \( \sin\theta\cos\alpha+\cos\theta\sin\alpha=\sin(\theta+\alpha) \).
2. The function is \( \sin^{-1}x+\alpha \) (valid in the stated range), so the derivative is \( \frac{1}{\sqrt{1-x^2}} \).

Teacher's Note:
1. Recognise the sine addition formula with the 5-12-13 triangle.
2. The range ensures theta + alpha stays in the principal interval.

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

Explanation:
1. For \( 0<x<1 \), \( \sin^{-1}\sqrt{1-x^2}=\cos^{-1}x \).
2. The sum is \( 2\cos^{-1}x \), with derivative \( \frac{-2}{\sqrt{1-x^2}} \).

Teacher's Note:
1. Put x = cos theta to see the equality.
2. The two terms are equal, not complementary.

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

Explanation:
1. \( \tanh^{-1}\frac{2x}{1+x^2}=2\tanh^{-1}x \) since \( \tanh2u=\frac{2\tanh u}{1+\tanh^2u} \).
2. Derivative: \( \frac{2}{1-x^2} \).

Teacher's Note:
1. The double-angle formula for tanh is the hyperbolic analogue of tan 2A.
2. The formula holds only for |x| < 1.