JEE Mathematics Methods of Differentiation MCQs Set 03

Answer: (a) 1

Explanation:
1. Put \( x=\tan\theta \) with \( 0<\theta<\frac\pi4 \).
2. \( \cos^{-1}(\cos2\theta)=2\theta \) and \( \tan^{-1}(\tan2\theta)=2\theta \).
3. Both functions equal \( 2\tan^{-1}x \), so the ratio of derivatives is \( 1 \).

Teacher's Note:
1. When two functions are identical on an interval, the derivative of one with respect to the other is 1.
2. Check the interval so that both principal-value reductions are valid.

Answer: (b) 1

Explanation:
1. Put \( x=\sin\theta \) with \( 0<\theta<\frac\pi2 \).
2. \( \cos^{-1}\sqrt{1-x^2}=\cos^{-1}(\cos\theta)=\theta=\sin^{-1}x \).
3. The two functions are equal, so the derivative is \( 1 \).

Teacher's Note:
1. On the stated interval the two functions coincide.
2. Substitution x = sin theta shows this at once.

Answer: (c) \( -1 \)

Explanation:
1. Differentiate: \( 1+y'=\cos(x+y)(1+y') \).
2. So \( (1+y')(1-\cos(x+y))=0 \).
3. Away from points where \( \cos(x+y)=1 \) this gives \( y'=-1 \).

Teacher's Note:
1. Factor out (1 + y') rather than dividing by it blindly.
2. The only real solution of t = sin t is t = 0, so x + y = 0 and y' = -1 directly.

Answer: (d) \( -\mathrm{cosec}\,x\cot x \)

Explanation:
1. \( e^{\log\mathrm{cosec}\,x}=\mathrm{cosec}\,x \) (in the interval where it is positive).
2. Its derivative is \( -\mathrm{cosec}\,x\cot x \).

Teacher's Note:
1. Exponential and logarithm cancel when they are inverse of each other.
2. Simplify before differentiating.

Answer: (a) \( \frac{\cos\sqrt{2x-3}}{\sqrt{2x-3}} \)

Explanation:
1. \( \cos\sqrt{2x-3}\cdot\frac{1}{2\sqrt{2x-3}}\cdot2 \).
2. The factors of 2 cancel, giving \( \frac{\cos\sqrt{2x-3}}{\sqrt{2x-3}} \).

Teacher's Note:
1. Three-layer chain rule: sine, root, linear.
2. The derivative of 2x - 3 is 2, which cancels the 2 from the root.

Answer: (d) \( x^{n-1}(1+n\log x) \)

Explanation:
1. Product rule: \( nx^{n-1}\log x+x^n\cdot\frac1x \).
2. Factor \( x^{n-1} \): \( x^{n-1}(1+n\log x) \).

Teacher's Note:
1. Factor out the lowest power of x to present the answer neatly.
2. The derivative of log x contributes x^{n-1}.

Answer: (d) \( -\frac12 \)

Explanation:
1. \( \frac{\cos x}{1+\sin x}=\frac{\sin(\frac\pi2-x)}{1+\cos(\frac\pi2-x)}=\tan\left(\frac\pi4-\frac x2\right) \).
2. So \( y=\frac\pi4-\frac x2 \) and \( y'=-\frac12 \).

Teacher's Note:
1. Write cos x and sin x in terms of pi/2 - x to see the half-angle tangent.
2. The result is a linear function of x.

Answer: (a) 0

Explanation:
1. For \( x>0 \), \( \tan^{-1}u+\tan^{-1}\frac1u=\frac\pi2 \) with \( u=\frac{x}{1+x^2} \).
2. The sum is the constant \( \frac\pi2 \), so its derivative is \( 0 \).

Teacher's Note:
1. The identity tan inverse u plus tan inverse 1/u equals pi/2 holds only for positive u.
2. A constant has zero derivative.

Answer: (d) \( y(\log2)^2\,2^x \)

Explanation:
1. \( \log y=2^x\log2 \).
2. \( \frac{y'}{y}=2^x\log2\cdot\log2 \), so \( y'=y(\log2)^2\,2^x \).

Teacher's Note:
1. A power tower needs logs applied level by level.
2. Each exponential level contributes a factor log 2.

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

Explanation:
1. Take logs: \( m\log x+n\log y=(m+n)\log(x+y) \).
2. Differentiate: \( \frac mx+\frac{ny'}{y}=\frac{(m+n)(1+y')}{x+y} \).
3. Solving for \( y' \) gives \( y'=\frac yx \).

Teacher's Note:
1. Both sides are homogeneous of degree m + n.
2. Hence y is proportional to x and dy/dx = y/x.

Answer: (d) \( x^{\cos x}\left(\frac{\cos x}{x}-\sin x\log x\right) \)

Explanation:
1. \( \log y=\cos x\log x \).
2. \( \frac{y'}{y}=\frac{\cos x}{x}-\sin x\log x \).

Teacher's Note:
1. Differentiating cos x log x gives two terms by the product rule.
2. The derivative of cos x brings a minus sign.

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

Explanation:
1. \( \frac{1}{\sqrt{1+(\sqrt x)^2}}\cdot\frac{1}{2\sqrt x} \).

Teacher's Note:
1. Chain rule with the inner function sqrt x.
2. Its derivative is 1/(2 sqrt x).

Answer: (c) \( -\mathrm{sech}(\log x)\tanh(\log x)\cdot\frac1x \)

Explanation:
1. \( \frac{d}{du}\mathrm{sech}\,u=-\mathrm{sech}\,u\tanh u \) with \( u=\log x \), \( u'=\frac1x \).

Teacher's Note:
1. The derivative of sech carries a minus sign and tanh.
2. Multiply by 1/x.

Answer: (b) \( \frac1t \)

Explanation:
1. \( \frac{dy}{dt}=2a \), \( \frac{dx}{dt}=2at \).
2. \( \frac{dy}{dx}=\frac{2a}{2at}=\frac1t \).

Teacher's Note:
1. This is the standard parametrisation of a parabola.
2. Divide the two derivatives with respect to t.

Answer: (b) \( \frac{x\,e^{\sin^{-1}x}}{\sqrt{1-x^2}} \)

Explanation:
1. \( \frac{d}{dx}e^{\sin^{-1}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} \) and \( \frac{d}{dx}\log x=\frac1x \).
2. Dividing gives \( \frac{x\,e^{\sin^{-1}x}}{\sqrt{1-x^2}} \).

Teacher's Note:
1. Dividing by 1/x is the same as multiplying by x.
2. Take care not to forget it.

Answer: (a) \( \frac12 \)

Explanation:
1. Put \( x=\tan\theta \): \( \frac{\sec\theta-1}{\tan\theta}=\tan\frac\theta2 \).
2. So the function is \( \frac\theta2=\frac12\tan^{-1}x \).
3. Its derivative with respect to \( \tan^{-1}x \) is \( \frac12 \).

Teacher's Note:
1. The expression is the half-angle tangent.
2. A constant multiple of tan inverse x differentiates to that multiple.

Answer: (b) 2

Explanation:
1. Put \( x=\cos\theta \), \( 0<\theta<\frac\pi2 \): \( \cos^{-1}(2\cos^2\theta-1)=\cos^{-1}(\cos2\theta)=2\theta \).
2. So the function is \( 2\cos^{-1}x \) and the required derivative is \( 2 \).

Teacher's Note:
1. The identity 2cos^2 - 1 = cos 2 theta is the key.
2. The interval makes 2 theta lie in [0, pi].

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

Explanation:
1. Differentiate: \( e^{x+y}(1+y')=y+xy' \).
2. Replace \( e^{x+y} \) by \( xy \): \( xy+xyy'=y+xy' \).
3. \( y'(xy-x)=y-xy \), so \( y'=\frac{y(1-x)}{x(y-1)} \).

Teacher's Note:
1. Use the original equation to substitute e^(x+y) after differentiating.
2. Group the y' terms and factor.

Answer: (c) \( 5\sin^4x\sin6x \)

Explanation:
1. Product rule: \( 5\sin^4x\cos x\sin5x+\sin^5x\cdot5\cos5x \).
2. Factor \( 5\sin^4x \): \( \cos x\sin5x+\sin x\cos5x=\sin6x \).

Teacher's Note:
1. Look for a compound-angle identity after factoring.
2. Here sin A cos B + cos A sin B collapses to sin 6x.

Answer: (c) \( -2 \)

Explanation:
1. \( \frac{1-\sin2x}{1+\sin2x}=\tan^2\left(\frac\pi4-x\right) \), so near \( x=0 \), \( y=\tan\left(\frac\pi4-x\right) \).
2. \( y'=-\sec^2\left(\frac\pi4-x\right) \); at \( x=0 \) this is \( -2 \).

Teacher's Note:
1. Write 1 - sin 2x as (cos x - sin x)^2 and 1 + sin 2x as (cos x + sin x)^2.
2. The tangent is positive near x = 0, so the root removes the square cleanly.

Answer: (b) \( \frac{-e^x}{(1+e^x)^3} \)

Explanation:
1. \( \frac{dx}{dy}=\frac{1}{1+e^x} \).
2. \( \frac{d^2x}{dy^2}=\frac{d}{dx}\left(\frac{1}{1+e^x}\right)\cdot\frac{dx}{dy}=\frac{-e^x}{(1+e^x)^2}\cdot\frac{1}{1+e^x} \).

Teacher's Note:
1. Differentiate with respect to y by multiplying the x-derivative by dx/dy.
2. That extra factor gives the cube.

Answer: (a) 1

Explanation:
1. At \( x=0 \): \( 0=y \), so the point is \( (0,0) \).
2. Differentiate: \( e^{xy}+xe^{xy}(y+xy')=y'+\sin2x \).
3. At \( (0,0) \): \( 1=y' \).

Teacher's Note:
1. First find the y value at the given x.
2. Then differentiate implicitly and substitute.

Answer: (a) \( -\frac{1}{10}\log2 \)

Explanation:
1. \( \frac{2^x}{1+2^{2x+1}}=\frac{2^{x+1}-2^x}{1+2^{x+1}\cdot2^x} \), so \( y=\tan^{-1}2^{x+1}-\tan^{-1}2^x \).
2. \( y'=\frac{2^{x+1}\log2}{1+2^{2x+2}}-\frac{2^x\log2}{1+2^{2x}} \).
3. At \( x=0 \): \( \frac{2\log2}{5}-\frac{\log2}{2}=-\frac{\log2}{10} \).

Teacher's Note:
1. Use tan inverse a - tan inverse b = tan inverse of (a-b)/(1+ab).
2. Evaluate each fraction separately at x = 0.

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

Explanation:
1. Put \( x=a\sin\theta \): \( \tan^{-1}(\tan\theta)=\theta=\sin^{-1}\frac xa \).
2. Derivative: \( \frac{1}{\sqrt{a^2-x^2}} \).

Teacher's Note:
1. Whenever sqrt(a^2 - x^2) appears, try x = a sin theta.
2. The answer matches the derivative of sin inverse of x/a.