JEE Mathematics Differentiation MCQs Set 08

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. Put \( x^2=\cos2\theta \): \( \sqrt{\dfrac{1-x^2}{1+x^2}}=\tan\theta \), so the first function is \( \theta=\dfrac12\cos^{-1}x^2 \).
2. Hence it is half of \( \cos^{-1}(x^2) \) and the required derivative is \( \dfrac12 \).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

Answer: (c) \( \frac{1}{4} \)

Explanation:
1. Put \( x=\tan\theta \): \( \dfrac{\sec\theta-1}{\tan\theta}=\tan\dfrac\theta2 \), so \( u=\dfrac\theta2=\dfrac12\tan^{-1}x \).
2. \( v=2\tan^{-1}x \), so \( \dfrac{du}{dv}=\dfrac{1/2}{2}=\dfrac14 \).

Teacher's Note:
1. The derivative of u with respect to v is (du/dx) divided by (dv/dx).
2. Simplify both functions with a suitable substitution to get a constant ratio quickly.

Answer: (b) \( 1 \)

Explanation:
1. For \( 0<x<1 \), \( \sin^{-1}\dfrac{2x}{1+x^2}=2\tan^{-1}x \) and \( \cos^{-1}\dfrac{1-x^2}{1+x^2}=2\tan^{-1}x \).
2. The two functions are equal, so the derivative of one with respect to the other is 1.

Teacher's Note:
1. Simplify both functions with a suitable substitution to get a constant ratio quickly.
2. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. Put \( x=\sin\theta \): \( \dfrac{\sin\theta}{1+\cos\theta}=\tan\dfrac\theta2 \), so the function is \( \dfrac\theta2=\dfrac12\sin^{-1}x \).
2. The derivative with respect to \( \sin^{-1}x \) is \( \dfrac12 \).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

Answer: (c) \( \frac{2}{3} \)

Explanation:
1. Put \( x=\tan\theta \): the first function is \( 2\theta \) and \( \dfrac{1-3x^2}{3x-x^3}=\cot3\theta \), so the second is \( 3\theta \).
2. The ratio of derivatives is \( \dfrac23 \).

Teacher's Note:
1. The derivative of u with respect to v is (du/dx) divided by (dv/dx).
2. Simplify both functions with a suitable substitution to get a constant ratio quickly.

Answer: (c) \( e^{\sin^{-1}{\left(x \right)}} \)

Explanation:
1. Let \( u=\sin^{-1}x \). Then \( \dfrac{d}{du}e^u=e^u=e^{\sin^{-1}x} \).

Teacher's Note:
1. Simplify both functions with a suitable substitution to get a constant ratio quickly.
2. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.

Answer: (c) \( \frac{1}{\left(\tan^{-1}{\left(x \right)} + 1\right)^{2}} \)

Explanation:
1. Let \( u=\tan^{-1}x \). Then \( \dfrac{d}{du}\dfrac u{1+u}=\dfrac{1}{(1+u)^2} \).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

Answer: (b) \( \frac{1}{4} \)

Explanation:
1. Put \( x=\tan\theta \) in the first function: it equals \( \dfrac\theta2=\dfrac12\tan^{-1}x \).
2. Put \( x=\sin\phi \) in the second: \( \dfrac{2\sin\phi\cos\phi}{1-2\sin^2\phi}=\tan2\phi \), so it equals \( 2\sin^{-1}x \) near \( x=0 \).
3. Derivatives at 0: \( \dfrac12\cdot1 \) and \( 2\cdot1 \), so the ratio is \( \dfrac{1/2}{2}=\dfrac14 \).

Teacher's Note:
1. The derivative of u with respect to v is (du/dx) divided by (dv/dx).
2. Simplify both functions with a suitable substitution to get a constant ratio quickly.

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

Explanation:
1. Put \( x=\cos\theta \): the first function is \( \tan^{-1}\tan\theta=\theta \) and the second is \( \cos^{-1}\sin2\theta=\dfrac\pi2-2\theta \) for small \( \theta \).
2. The ratio of derivatives is \( \dfrac{1}{-2}=-\dfrac12 \).

Teacher's Note:
1. Simplify both functions with a suitable substitution to get a constant ratio quickly.
2. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.

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

Explanation:
1. Put \( ax=\sin\theta \): the first function is \( 2\theta \) and \( \sqrt{1-a^2x^2}=\cos\theta \).
2. \( \dfrac{d(2\theta)}{d(\cos\theta)}=\dfrac{2}{-\sin\theta}=-\dfrac{2}{ax} \).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

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

Explanation:
1. \( \tan^{-1}\dfrac{1+ax}{1-ax}=\dfrac\pi4+\tan^{-1}(ax) \), whose derivative is \( \dfrac{a}{1+a^2x^2} \).
2. \( \dfrac{d}{dx}\sqrt{1+a^2x^2}=\dfrac{a^2x}{\sqrt{1+a^2x^2}} \).
3. Ratio: \( \dfrac{a}{1+a^2x^2}\cdot\dfrac{\sqrt{1+a^2x^2}}{a^2x}=\dfrac{1}{ax\sqrt{1+a^2x^2}} \).

Teacher's Note:
1. The derivative of u with respect to v is (du/dx) divided by (dv/dx).
2. Simplify both functions with a suitable substitution to get a constant ratio quickly.

Answer: (b) \( (2m)! \)

Explanation:
1. \( (x^2-1)^m \) is a polynomial of degree \( 2m \) with leading term \( x^{2m} \).
2. Differentiating \( 2m \) times kills all lower terms and leaves \( (2m)! \).

Teacher's Note:
1. Differentiate two or three times to see the pattern and then generalise.
2. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).

Answer: (c) \( x=-n \)

Explanation:
1. By Leibnitz's rule, \( \dfrac{d^n}{dx^n}(xe^x)=xe^x+ne^x=(x+n)e^x \).
2. This is zero when \( x=-n \).

Teacher's Note:
1. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).
2. Convert products of trigonometric functions to sums before taking the nth derivative.

Answer: (d) \( y \)

Explanation:
1. The series is that of \( e^{-x} \), so \( y=e^{-x} \).
2. \( y''=e^{-x}=y \).

Teacher's Note:
1. Substituting the original expression back into the differential equation is a good check.
2. Differentiate the given relation twice and eliminate the constants.

Answer: (d) \( -2^2(\cos2x+2^2\cos4x) \)

Explanation:
1. \( 2\cos x\cos3x=\cos4x+\cos2x \).
2. Differentiate twice: \( -16\cos4x-4\cos2x=-2^2(\cos2x+2^2\cos4x) \).

Teacher's Note:
1. Convert products of trigonometric functions to sums before taking the nth derivative.
2. Differentiate two or three times to see the pattern and then generalise.

Answer: (c) \( 0 \)

Explanation:
1. \( y''=m^2ae^{mx}+m^2be^{-mx}=m^2y \).
2. So \( y''-m^2y=0 \).

Teacher's Note:
1. Differentiate the given relation twice and eliminate the constants.
2. Find dy/dx first and then differentiate again, using the product or chain rule.

Answer: (d) \( f''(e^x)e^{2x}+f'(e^x)e^x \)

Explanation:
1. First derivative: \( f'(e^x)e^x \).
2. Second derivative by the product rule: \( f''(e^x)e^x\cdot e^x+f'(e^x)e^x=f''(e^x)e^{2x}+f'(e^x)e^x \).

Teacher's Note:
1. Write the composite as f(g(x)) first so no factor of g'(x) is forgotten.
2. Simplify the inside expression where possible before applying the chain rule.

Answer: (a) \( 0 \)

Explanation:
1. For small x, \( y=x^3\left(x-\dfrac{x^2}2+\cdots\right)=x^4-\dfrac{x^5}{2}+\cdots \).
2. The lowest power of x is \( x^4 \), so \( y''(0)=0 \).

Teacher's Note:
1. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).
2. Convert products of trigonometric functions to sums before taking the nth derivative.

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

Explanation:
1. \( \dfrac{dx}{dy}=\dfrac1{dy/dx} \).
2. \( \dfrac{d^2x}{dy^2}=\dfrac{d}{dx}\left(\dfrac1{y'}\right)\dfrac{dx}{dy}=-\dfrac{y''}{y'^2}\cdot\dfrac1{y'}=-\dfrac{y''}{y'^3} \).

Teacher's Note:
1. Differentiate the given relation twice and eliminate the constants.
2. Find dy/dx first and then differentiate again, using the product or chain rule.

Answer: (c) \( 2^n[e^{2x}+(-1)^ne^{-2x}] \)

Explanation:
1. \( \dfrac{d^n}{dx^n}e^{2x}=2^ne^{2x} \) and \( \dfrac{d^n}{dx^n}e^{-2x}=(-2)^ne^{-2x} \).
2. Adding gives \( 2^n[e^{2x}+(-1)^ne^{-2x}] \).

Teacher's Note:
1. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).
2. Convert products of trigonometric functions to sums before taking the nth derivative.

Answer: (b) \( \dfrac{n!}{(1-x)^{n+1}} \)

Explanation:
1. \( \dfrac x{1-x}=-1+\dfrac1{1-x} \).
2. \( \dfrac{d^n}{dx^n}(1-x)^{-1}=n!(1-x)^{-(n+1)} \) (each differentiation brings a factor \( +1 \)).

Teacher's Note:
1. Differentiate two or three times to see the pattern and then generalise.
2. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).

Answer: (a) \( 2^{2n-1}\sin\left(4x+\dfrac{n\pi}2\right) \)

Explanation:
1. \( \sin2x\cos2x=\dfrac12\sin4x \).
2. \( y_n=\dfrac12\cdot4^n\sin\left(4x+\dfrac{n\pi}2\right)=2^{2n-1}\sin\left(4x+\dfrac{n\pi}2\right) \).

Teacher's Note:
1. Convert products of trigonometric functions to sums before taking the nth derivative.
2. Differentiate two or three times to see the pattern and then generalise.

Answer: (b) \( (-5)^ne^{6-5x} \)

Explanation:
1. Each differentiation multiplies by \( -5 \).
2. So \( y_n=(-5)^ne^{6-5x} \).

Teacher's Note:
1. Differentiate two or three times to see the pattern and then generalise.
2. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).

Answer: (d) \( 8^x(\log_e8)^n \)

Explanation:
1. \( \dfrac{d}{dx}8^x=8^x\log_e8 \), and each further differentiation brings another factor \( \log_e8 \).
2. So \( y_n=8^x(\log_e8)^n \).

Teacher's Note:
1. Use the standard nth derivatives of sin(ax+b), cos(ax+b), e^(ax) and 1/(ax+b).
2. Convert products of trigonometric functions to sums before taking the nth derivative.

Answer: (b) \( a^nf_n(ax+b) \)

Explanation:
1. Each differentiation by the chain rule brings a factor a.
2. So \( D^n[f(ax+b)]=a^nf^{(n)}(ax+b) \).

Teacher's Note:
1. Convert products of trigonometric functions to sums before taking the nth derivative.
2. Differentiate two or three times to see the pattern and then generalise.