JEE Mathematics Methods of Differentiation MCQs Set 05

Answer: (c) \( (x^x)^x\,(1+2\log x)\,x \)

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

Teacher's Note:
1. Simplify the power tower using (a^b)^c = a^(bc).
2. Then use logarithmic differentiation.

Answer: (c) \( y\left\{\frac{a(cx+d)}{ax+b}+c\log(ax+b)\right\} \)

Explanation:
1. \( \log y=(cx+d)\log(ax+b) \).
2. \( \frac{y'}{y}=c\log(ax+b)+\frac{a(cx+d)}{ax+b} \).

Teacher's Note:
1. Product rule on (cx + d) and log(ax + b).
2. The chain factor a appears in the fraction.

Answer: (a) 1

Explanation:
1. Both \( x \) and \( y \) equal \( \tan^{-1}t \) (for \( t>0 \)).
2. So \( y=x \) and \( \frac{dy}{dx}=1 \).

Teacher's Note:
1. Two functions of t that simplify to the same expression give slope 1.
2. Use a right triangle with legs t and 1.

Answer: (c) \( \frac ba\tan\theta \)

Explanation:
1. Differentiate: \( \frac{dx}{d\theta}=2a[\cos2\theta+\cos4\theta] \) and \( \frac{dy}{d\theta}=2b[-\sin2\theta+\sin4\theta] \).
2. Using sum-to-product: \( \frac{dy}{dx}=\frac ba\cdot\frac{2\cos3\theta\sin\theta}{2\cos3\theta\cos\theta}=\frac ba\tan\theta \).

Teacher's Note:
1. Use sum-to-product to factor out cos 3 theta.
2. The common factor cancels, leaving tan theta.

Answer: (d) \( -1 \)

Explanation:
1. Square the first: \( x^4+y^4+2x^2y^2=t^2+\frac1{t^2}+2 \).
2. Subtract the second: \( 2x^2y^2=2 \), so \( x^2y^2=1 \), \( xy=\pm1 \) constant.
3. Hence \( y'=-\frac yx \) and \( x^3yy'=-x^2y^2=-1 \).

Teacher's Note:
1. Eliminate t by squaring.
2. A constant product xy means dy/dx = -y/x.

Answer: (b) 0

Explanation:
1. \( \sec^{-1}\frac{1}{2x^2-1}=\cos^{-1}(2x^2-1) \), whose derivative at \( x=-\frac13 \) is finite and non-zero: \( \frac{2}{\sqrt{1-x^2}}\cdot(\pm1)=\pm\frac{3}{\sqrt2} \).
2. \( \frac{d}{dx}\sqrt{1+3x}=\frac{3}{2\sqrt{1+3x}}\to\infty \) at \( x=-\frac13 \).
3. A finite number divided by an infinite one is \( 0 \).

Teacher's Note:
1. Compute each derivative separately and examine the limit.
2. The denominator derivative blows up at the given point.

Answer: (b) \( \frac{y^2-x}{2y^3-2xy-1} \)

Explanation:
1. The pattern repeats, so the inner radical \( \sqrt{y+\cdots} \) equals \( \sqrt{y+y}=\sqrt{2y} \), hence \( y^2=x+\sqrt{2y} \).
2. Square again: \( (y^2-x)^2=2y \).
3. Differentiate: \( 2(y^2-x)(2yy'-1)=2y' \), so \( y'=\frac{y^2-x}{2y^3-2xy-1} \).

Teacher's Note:
1. Find the repeating unit and replace it by y.
2. Square twice and differentiate implicitly.

Answer: (c) \( \frac{-\sin^2y}{4} \)

Explanation:
1. \( x=3+4\cot y \).
2. \( \frac{dx}{dy}=-4\,\mathrm{cosec}^2y \).
3. \( \frac{dy}{dx}=-\frac{\sin^2y}{4} \).

Teacher's Note:
1. Solve for x as a function of y when that is easy.
2. Take the reciprocal of dx/dy.

Answer: (d) 0

Explanation:
1. \( dy=-\frac{dx}{x^2} \) and \( \sqrt{1+y^4}=\frac{\sqrt{x^4+1}}{x^2} \).
2. \( \frac{dy}{\sqrt{1+y^4}}=-\frac{dx}{\sqrt{1+x^4}} \).
3. So the sum is \( 0 \).

Teacher's Note:
1. Work with differentials and substitute y = 1/x.
2. The terms cancel exactly.

Answer: (b) \( 2+(g(x))^n \)

Explanation:
1. Since \( f(g(x))=x \), differentiating gives \( f'(g(x))\,g'(x)=1 \).
2. \( f'(g(x))=\frac{1}{2+(g(x))^n} \), so \( g'(x)=2+(g(x))^n \).

Teacher's Note:
1. The derivative of an inverse function is the reciprocal of f' evaluated at g(x).
2. Substitute g(x) into f', not x.

Answer: (d) \( \frac{f'(x)\log f(x)}{g'(y)\left(1+\log f(x)\right)^2} \)

Explanation:
1. Take logs: \( g(y)\log f=f-g(y) \), so \( g(y)=\frac{f}{1+\log f} \).
2. Differentiate: \( g'(y)y'=\frac{f'(1+\log f)-f\cdot\frac{f'}{f}}{(1+\log f)^2}=\frac{f'\log f}{(1+\log f)^2} \).
3. Divide by \( g'(y) \).

Teacher's Note:
1. Isolate g(y) first, then differentiate once.
2. The numerator simplifies because f times f'/f equals f'.

Answer: (a) \( g(a) \)

Explanation:
1. By definition \( f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}g(x) \).
2. Continuity of \( g \) at \( a \) gives \( g(a) \).

Teacher's Note:
1. We cannot use the product rule because g may not be differentiable.
2. Go back to the limit definition.

Answer: (c) a constant

Explanation:
1. Divide by \( |x-y| \): \( \left|\frac{f(x)-f(y)}{x-y}\right|\le|x-y|^2 \).
2. Letting \( y\to x \) gives \( f'(x)=0 \) for all \( x \).
3. So \( f \) is constant.

Teacher's Note:
1. An exponent greater than 1 on the right side forces the derivative to be zero.
2. Zero derivative everywhere means constant.

Answer: (a) \( \frac{\phi''}{\phi}-\frac{f''}{f}-\frac{g''}{g} \)

Explanation:
1. \( \phi''=f''g+2f'g'+fg''=f''g+2k+fg'' \).
2. Divide by \( \phi=fg \): \( \frac{\phi''}{\phi}=\frac{f''}{f}+\frac{2k}{fg}+\frac{g''}{g} \).
3. Rearranging gives option (a).

Teacher's Note:
1. Use the second-order product rule (Leibniz rule).
2. Divide by fg to separate the terms.

Answer: (d) \( \frac{-(\sinh^3x+\cosh^3x)}{\cosh^2x\sinh^2x} \)

Explanation:
1. \( e^{\frac12\log(\mathrm{sech}^2x)}=\mathrm{sech}\,x \) and \( 3^{\frac12\log_3(\mathrm{cosech}^2x)}=\mathrm{cosech}\,x \) (for \( x>0 \)).
2. Derivative: \( -\mathrm{sech}\,x\tanh x-\mathrm{cosech}\,x\coth x=-\frac{\sinh x}{\cosh^2x}-\frac{\cosh x}{\sinh^2x} \).
3. Common denominator gives the answer.

Teacher's Note:
1. Use 1 - tanh^2 = sech^2 and coth^2 - 1 = cosech^2.
2. Then differentiate the two simple hyperbolic functions.

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

Explanation:
1. Note that \( (1+x)(1+x^2)(1+x^4)\cdots=\frac{1}{1-x} \).
2. The terms are the logarithmic derivative of this product.
3. Hence the sum is \( \frac{d}{dx}\log\frac{1}{1-x}=\frac{1}{1-x} \).

Teacher's Note:
1. Recognise a sum of terms of the form u'/u as a log derivative of a product.
2. The infinite product telescopes using the difference of squares.

Answer: (a) 5

Explanation:
1. With \( t=\tan x \): \( f=(1+t)\left(1+\frac{1-t}{1+t}\right)=(1+t)+(1-t)=2 \).
2. \( g'(x)=2x+1 \), so \( g'[f(x)]=g'(2)=5 \).

Teacher's Note:
1. The function f is secretly constant.
2. Evaluate g' at that constant.

Answer: (d) \( -25y \)

Explanation:
1. \( \frac{dy}{dx}=\frac{5\cos5\theta}{-\sin\theta} \).
2. \( \frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\frac{-5\cos5\theta}{\sin\theta}\right)\cdot\frac{1}{-\sin\theta} \), which simplifies to \( (1-x^2)y''-xy'=-25y \).

Teacher's Note:
1. This is the Chebyshev-type equation with n = 5.
2. The same relation gives -n^2 y for y = sin n theta.

Answer: (d) d, c, b, a

Explanation:
1. \( P(0)=d=4 \), \( P'(0)=c=3 \).
2. \( P''(0)=2b=4 \), so \( b=2 \).
3. \( P'''(0)=6a=6 \), so \( a=1 \).
4. Descending order of the values \( 4,3,2,1 \) is \( d,c,b,a \).

Teacher's Note:
1. Derivatives at 0 pick out the coefficients times factorials.
2. Compare the four values to order them.

Answer: (c) 2

Explanation:
1. The condition is \( n(a+b)^{n-1}=na^{n-1}+nb^{n-1} \).
2. For \( n=2 \): \( 2(a+b)=2a+2b \) holds.
3. For \( n=1 \) it gives \( 1=2 \) and for \( n\ge3 \) the identity fails (e.g. \( a=b=1 \) gives \( 2^{n-1}=2 \) only for \( n=2 \)).

Teacher's Note:
1. Test the condition at a = b = 1 to find n.
2. The relation is additive only for a linear derivative.

Answer: (c) \( (2,-1) \)

Explanation:
1. \( 1+x^2+x^4=(x^2+x+1)(x^2-x+1) \).
2. So the fraction equals \( x^2-x+1 \), with derivative \( 2x-1 \).
3. Hence \( (a,b)=(2,-1) \).

Teacher's Note:
1. Factor the numerator using the identity x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1).
2. Then differentiate the polynomial.

Answer: (b) \( -\tanh x \)

Explanation:
1. \( \frac{e^x-e^{-x}}{2}=\sinh x \) and \( \cos(\tan^{-1}\sinh x)=\frac{1}{\sqrt{1+\sinh^2x}}=\mathrm{sech}\,x \).
2. \( y=\log\mathrm{sech}\,x=-\log\cosh x \).
3. \( y'=-\tanh x \).

Teacher's Note:
1. Use cos(tan inverse u) = 1 over root(1 + u^2).
2. Then the logarithm of sech is minus log cosh.

Answer: (b) \( \frac{e^x}{\sqrt{e^{2x}+4e^x+5}} \)

Explanation:
1. With \( u=e^x+2 \), \( e^{2x}+4e^x+5=u^2+1 \).
2. \( \frac{d}{du}\log(u+\sqrt{u^2+1})=\frac{1}{\sqrt{u^2+1}} \).
3. Multiply by \( u'=e^x \).

Teacher's Note:
1. Complete the square under the root.
2. The expression is the inverse hyperbolic sine of u.

Answer: (b) \( \frac{x^2}{1-x^4} \)

Explanation:
1. \( y'=\frac14\left(\frac{1}{1+x}+\frac{1}{1-x}\right)-\frac{1}{2(1+x^2)}=\frac{1}{2(1-x^2)}-\frac{1}{2(1+x^2)} \).
2. Combine: \( \frac{(1+x^2)-(1-x^2)}{2(1-x^4)}=\frac{x^2}{1-x^4} \).

Teacher's Note:
1. Expand the log of a quotient into a difference first.
2. A common denominator 1 - x^4 appears.