JEE Mathematics Differentiation MCQs Set 04

Answer: (d) Does not exist

Explanation:
1. The function is continuous at 0: both sides tend to 1.
2. Left derivative at 0 is the derivative of the constant 1, which is 0.
3. Right derivative at 0 is \( \cos 0=1 \).
4. The two one-sided derivatives differ, so \( f'(0) \) does not exist.

Teacher's Note:
1. Remove the modulus using the sign of the expression near the point, then differentiate.
2. At a corner the left and right derivatives differ, so the derivative does not exist.

Answer: (d) Does not exist

Explanation:
1. Near 0, \( y=1-x \) for \( x\ge0 \) and \( y=1+x \) for \( x<0 \).
2. Right derivative is \( -1 \) and left derivative is \( +1 \).
3. They are unequal, so the derivative does not exist.

Teacher's Note:
1. At a corner the left and right derivatives differ, so the derivative does not exist.
2. Check one-sided derivatives wherever the expression inside the modulus vanishes.

Answer: (b) \( 1 \)

Explanation:
1. \( f'(x)=1 \) for every x.
2. So \( f'(f(x))=1 \) at every point, including \( x=4 \).

Teacher's Note:
1. Use the given values of the function and its derivative at the point after applying the rule.
2. Treat f as an unknown function and apply the chain rule symbolically.

Answer: (a) \( af'(a)-f(a) \)

Explanation:
1. Add and subtract \( af(a) \) in the numerator: \( af(x)-xf(a)=a[f(x)-f(a)]-f(a)(x-a) \).
2. Dividing by \( x-a \) and letting \( x\to a \) gives \( af'(a)-f(a) \).

Teacher's Note:
1. Treat f as an unknown function and apply the chain rule symbolically.
2. Substitute the point only after you have found the derivative expression.

Answer: (d) Does not exist

Explanation:
1. For \( x>3 \), \( f(x)=x-3 \) with slope 1; for \( x<3 \), \( f(x)=3-x \) with slope \( -1 \).
2. The one-sided derivatives at 3 are unequal, so \( f'(3) \) does not exist.

Teacher's Note:
1. Check one-sided derivatives wherever the expression inside the modulus vanishes.
2. Remove the modulus using the sign of the expression near the point, then differentiate.

Answer: (a) \( e^x\left[\log(1+x^2)+\dfrac{2x}{1+x^2}\right] \)

Explanation:
1. Product rule: \( e^x\log(1+x^2)+e^x\cdot\dfrac{2x}{1+x^2} \).
2. Taking \( e^x \) common gives option (a).

Teacher's Note:
1. Use the product rule (first times derivative of second plus second times derivative of first).
2. Check each term of the product rule separately before combining them.

Answer: (b) \( 3 x^{2} e^{x^{3}} \)

Explanation:
1. Chain rule: \( e^{x^3}\cdot\dfrac{d}{dx}(x^3) \).
2. \( =3x^2e^{x^3} \).

Teacher's Note:
1. The derivative of e^u is e^u times u', and a^u gives a^u log a times u'.
2. Differentiate the exponent first and then multiply by the original exponential.

Answer: (c) \( \left(\log{\left(x \right)} + \frac{1}{x}\right) e^{x} \)

Explanation:
1. \( y'=e^x\log x+e^x\cdot\dfrac1x \).
2. \( =e^x\left(\dfrac1x+\log x\right) \).

Teacher's Note:
1. Keep the common exponential or trigonometric factor outside the bracket to compare with the options quickly.
2. Use the product rule (first times derivative of second plus second times derivative of first).

Answer: (c) \( 2 \)

Explanation:
1. \( \dfrac{d}{dx}\log f(u)=\dfrac{f'(u)}{f(u)}\cdot u' \) with \( u=e^x+2x \).
2. At \( x=0 \): \( u=1 \) and \( u'=e^0+2=3 \).
3. Value \( =\dfrac{2}{3}\cdot3=2 \).

Teacher's Note:
1. Substitute the point only after you have found the derivative expression.
2. Use the given values of the function and its derivative at the point after applying the rule.

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

Explanation:
1. \( y'=\dfrac{1}{\sqrt{1-x}}\cdot\dfrac{1}{2\sqrt x} \).
2. \( =\dfrac{1}{2\sqrt x\sqrt{1-x}} \).

Teacher's Note:
1. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.
2. State the interval on which the simplification is valid, since it decides the sign of the answer.

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

Explanation:
1. For \( 0<x<1 \), \( \sin^{-1}\sqrt{1-x^2}=\cos^{-1}x \).
2. So \( \dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-x^2}} \).

Teacher's Note:
1. State the interval on which the simplification is valid, since it decides the sign of the answer.
2. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.

Answer: (d) \( - \frac{1}{4} \)

Explanation:
1. Let \( \tan A=\sqrt x \) and \( \tan B=x \). Then \( \tan(A-B)=\dfrac{\sqrt x-x}{1+x^{3/2}} \).
2. So \( y=\tan^{-1}\sqrt x-\tan^{-1}x \) and \( y'=\dfrac{1}{2\sqrt x(1+x)}-\dfrac1{1+x^2} \).
3. At \( x=1 \): \( \dfrac14-\dfrac12=-\dfrac14 \).

Teacher's Note:
1. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.
2. State the interval on which the simplification is valid, since it decides the sign of the answer.

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

Explanation:
1. Standard result for \( |x|>1 \), \( x>1 \): \( \dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^2-1}} \).

Teacher's Note:
1. State the interval on which the simplification is valid, since it decides the sign of the answer.
2. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.

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

Explanation:
1. \( \cot^{-1}u=\dfrac\pi2-\tan^{-1}u \) and \( \tan^{-1}\dfrac{1+x}{1-x}=\dfrac\pi4+\tan^{-1}x \).
2. So \( y=\dfrac\pi4-\tan^{-1}x \) and \( y'=-\dfrac{1}{1+x^2} \).

Teacher's Note:
1. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.
2. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.

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

Explanation:
1. \( \sqrt{\dfrac{1+\cos x}{1-\cos x}}=\cot\dfrac x2 \) for \( 0<x<\pi \).
2. So \( y=\tan^{-1}\cot\dfrac x2=\dfrac\pi2-\dfrac x2 \), giving \( y'=-\dfrac12 \).

Teacher's Note:
1. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.
2. State the interval on which the simplification is valid, since it decides the sign of the answer.

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

Explanation:
1. Put \( x=\sin\theta \): \( y=\tan^{-1}\tan\theta=\theta=\sin^{-1}x \).
2. So \( y'=\dfrac1{\sqrt{1-x^2}} \).

Teacher's Note:
1. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.
2. State the interval on which the simplification is valid, since it decides the sign of the answer.

Answer: (c) \( \pm\dfrac{2}{1+x^2} \)

Explanation:
1. Put \( x=\tan\theta \): \( \dfrac{1-x^2}{1+x^2}=\cos2\theta \), so \( y=\sin^{-1}\cos2\theta=\dfrac\pi2-2\theta \) when \( x>0 \) (small).
2. Then \( y=\dfrac\pi2-2\tan^{-1}x \) for \( x>0 \) and \( y=\dfrac\pi2+2\tan^{-1}x \) for \( x<0 \).
3. So \( \dfrac{dy}{dx}=\pm\dfrac{2}{1+x^2} \), the sign depending on the sign of x.

Teacher's Note:
1. State the interval on which the simplification is valid, since it decides the sign of the answer.
2. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.

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

Explanation:
1. \( \tan^{-1}\dfrac{a-x}{1+ax}=\tan^{-1}a-\tan^{-1}x \), because a is a constant.
2. Differentiating gives \( 0-\dfrac1{1+x^2} \).

Teacher's Note:
1. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.
2. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.

Answer: (d) \( 1 \)

Explanation:
1. Divide numerator and denominator by \( \cos x \): the argument is \( \dfrac{\tan x+1}{1-\tan x}=\tan\left(\dfrac\pi4+x\right) \).
2. So \( y=\dfrac\pi4+x \) and \( \dfrac{dy}{dx}=1 \).

Teacher's Note:
1. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.
2. State the interval on which the simplification is valid, since it decides the sign of the answer.

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

Explanation:
1. \( \dfrac{\cos x}{1+\sin x}=\tan\left(\dfrac\pi4-\dfrac x2\right) \).
2. So the function equals \( \dfrac\pi4-\dfrac x2 \) and its derivative is \( -\dfrac12 \).

Teacher's Note:
1. State the interval on which the simplification is valid, since it decides the sign of the answer.
2. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.

Answer: (c) \( \frac{4}{x^{2} + 1} \)

Explanation:
1. Put \( x=\tan\theta \) with \( |x|<1 \). Then \( \sin^{-1}\dfrac{2x}{1+x^2}=2\theta \) and \( \sec^{-1}\dfrac{1+x^2}{1-x^2}=\cos^{-1}(\cos2\theta)=2\theta \).
2. So \( y=4\tan^{-1}x \) and \( y'=\dfrac4{1+x^2} \).

Teacher's Note:
1. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.
2. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.

Answer: (c) \( \frac{5}{25 x^{2} + 1} \)

Explanation:
1. \( \dfrac{4x}{1+5x^2}=\dfrac{5x-x}{1+5x\cdot x} \), so the first term is \( \tan^{-1}5x-\tan^{-1}x \).
2. \( \dfrac{2+3x}{3-2x}=\dfrac{\frac23+x}{1-\frac23x} \), so the second term is \( \tan^{-1}\dfrac23+\tan^{-1}x \).
3. Adding: \( y=\tan^{-1}5x+\text{constant} \) and \( y'=\dfrac{5}{1+25x^2} \).

Teacher's Note:
1. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.
2. State the interval on which the simplification is valid, since it decides the sign of the answer.

Answer: (a) \( (1-x^2)\dfrac{dy}{dx}-xy-1=0 \)

Explanation:
1. \( y'=\dfrac{\frac{1}{\sqrt{1-x^2}}\sqrt{1-x^2}-\sin^{-1}x\cdot\frac{-x}{\sqrt{1-x^2}}}{1-x^2}=\dfrac{1+xy}{1-x^2} \).
2. So \( (1-x^2)y'-xy-1=0 \).

Teacher's Note:
1. Work from the outside in and multiply by the derivative of the inner function at every layer.
2. Write the composite as f(g(x)) first so no factor of g'(x) is forgotten.

Answer: (a) \( \frac{3}{\sqrt{1 - x^{2}}} \)

Explanation:
1. Put \( x=\sin\theta \): \( 3x-4x^3=\sin3\theta \).
2. Near \( x=0 \), \( y=3\sin^{-1}x \), so \( y'=\dfrac{3}{\sqrt{1-x^2}} \).

Teacher's Note:
1. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.
2. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.

Answer: (a) \( 1 \)

Explanation:
1. \( \dfrac{\sin x+\cos x}{\sqrt2}=\sin\left(x+\dfrac\pi4\right) \).
2. So \( y=x+\dfrac\pi4 \) near \( x=0 \) and \( \dfrac{dy}{dx}=1 \).

Teacher's Note:
1. Substitute x = tan t, sin t or cos t to collapse the inverse trigonometric expression into a simple multiple of t.
2. Know the standard forms such as sin^-1(2x/(1+x^2)) = 2 tan^-1 x before differentiating.