JEE Mathematics Differentiation MCQs Set 03

Answer: (a) \( n\sin^{n-1}x\cos(n+1)x \)

Explanation:
1. Product rule: \( n\sin^{n-1}x\cos x\cos nx-n\sin^nx\sin nx \).
2. Factor \( n\sin^{n-1}x \): \( \cos x\cos nx-\sin x\sin nx=\cos(n+1)x \).
3. Result: \( n\sin^{n-1}x\cos(n+1)x \).

Teacher's Note:
1. Use \( \cos(A+B)=\cos A\cos B-\sin A\sin B \) to combine the terms.
2. Check with \( n=1 \): the derivative of \( \sin x\cos x \) is \( \cos2x \).

Answer: (b) \( -\sin(\sin x^2)\cos x^2\cdot2x \)

Explanation:
1. Outer function: \( -\sin(\sin x^2) \).
2. Middle: \( \cos x^2 \). Inner: \( 2x \).
3. Product: \( -\sin(\sin x^2)\cos x^2\cdot2x \).

Teacher's Note:
1. Differentiate from the outside in, multiplying the three factors.
2. \( \cos x^2 \) is not the same as \( \cos^2x \).

Answer: (c) \( \dfrac12\dfrac{\cos\sqrt{\sin x+\cos x}}{\sqrt{\sin x+\cos x}}(\cos x-\sin x) \)

Explanation:
1. \( y'=\cos\sqrt{\sin x+\cos x}\cdot\dfrac{d}{dx}\sqrt{\sin x+\cos x} \).
2. \( \dfrac d{dx}\sqrt{u}=\dfrac{u'}{2\sqrt u} \) with \( u'=\cos x-\sin x \).
3. So \( y'=\dfrac{\cos\sqrt{\sin x+\cos x}}{2\sqrt{\sin x+\cos x}}(\cos x-\sin x) \).

Teacher's Note:
1. Do not forget the derivative of the expression inside the square root.
2. The factor \( \cos x-\sin x \) appears in option (c) only.

Answer: (d) \( \dfrac{4x}{(1-x^2)^2}\cos\left(\dfrac{1+x^2}{1-x^2}\right) \)

Explanation:
1. Chain rule: \( y'=\cos\left(\dfrac{1+x^2}{1-x^2}\right)\cdot\dfrac d{dx}\dfrac{1+x^2}{1-x^2} \).
2. \( \dfrac d{dx}\dfrac{1+x^2}{1-x^2}=\dfrac{2x(1-x^2)+2x(1+x^2)}{(1-x^2)^2}=\dfrac{4x}{(1-x^2)^2} \).
3. Hence option (d).

Teacher's Note:
1. Use the quotient rule for the inner function.
2. Simplify the numerator \( 2x-2x^3+2x+2x^3=4x \).

Answer: (c) \( 4(x^2+\cos x)^3(2x-\sin x) \)

Explanation:
1. Power rule with the chain rule: \( 4(x^2+\cos x)^3\cdot\dfrac{d}{dx}(x^2+\cos x) \).
2. \( \dfrac d{dx}(x^2+\cos x)=2x-\sin x \).
3. Result: \( 4(x^2+\cos x)^3(2x-\sin x) \).

Teacher's Note:
1. The exponent drops by 1, not to zero.
2. The derivative of \( \cos x \) is \( -\sin x \).

Answer: (d) \( -2\sin2x \)

Explanation:
1. \( \dfrac{\cot^2x-1}{\cot^2x+1}=\dfrac{\cot^2x-1}{\csc^2x}=\cos^2x-\sin^2x=\cos2x \).
2. Derivative of \( \cos2x \) is \( -2\sin2x \).

Teacher's Note:
1. Use \( 1+\cot^2x=\csc^2x \) to simplify.
2. Simplify before differentiating.

Answer: (d) \( -2\tan2x\sec2x \)

Explanation:
1. Multiply numerator and denominator by \( \sin x\cos x \): \( y=\dfrac{\sin^2x+\cos^2x}{\sin^2x-\cos^2x}=\dfrac1{-\cos2x}=-\sec2x \).
2. \( \dfrac{dy}{dx}=-2\sec2x\tan2x \).

Teacher's Note:
1. Express tangent and cotangent in sines and cosines.
2. The derivative of \( \sec2x \) is \( 2\sec2x\tan2x \).

Answer: (b) \( -4\csc2x\cot2x \)

Explanation:
1. \( \sec^2x+\csc^2x=\dfrac{1}{\sin^2x\cos^2x}=\dfrac4{\sin^22x} \).
2. The square root is \( 2\csc2x \) for \( \sin2x>0 \).
3. Derivative: \( 2\cdot(-2\csc2x\cot2x)=-4\csc2x\cot2x \).

Teacher's Note:
1. Combine \( \sec^2x+\csc^2x \) over a common denominator to obtain a perfect square.
2. The derivative of \( \csc2x \) is \( -2\csc2x\cot2x \).

Answer: (a) \( \dfrac{5(3-x)}{3(1-x)^{5/3}}-2\sin(4x+2) \)

Explanation:
1. First term: \( 5\left[(1-x)^{-2/3}+\dfrac23x(1-x)^{-5/3}\right]=\dfrac{5\left[(1-x)+\frac23x\right]}{(1-x)^{5/3}}=\dfrac{5(3-x)}{3(1-x)^{5/3}} \).
2. Second term: \( 2\cos(2x+1)\cdot(-\sin(2x+1))\cdot2=-2\sin(4x+2) \).
3. Sum as in option (a).

Teacher's Note:
1. Use the product rule on \( 5x(1-x)^{-2/3} \).
2. \( 2\sin\theta\cos\theta=\sin2\theta \) turns the second term into \( -2\sin(4x+2) \).

Answer: (b) \( \dfrac{\cos x}{2y-1} \)

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

Teacher's Note:
1. Square both sides to remove the root and differentiate implicitly.
2. Collect all terms in \( \frac{dy}{dx} \) on one side.

Answer: (a) \( \dfrac1x \)

Explanation:
1. For \( x>0 \): \( \dfrac d{dx}\log x=\dfrac1x \).
2. For \( x<0 \): \( \log(-x) \) has derivative \( \dfrac{-1}{-x}=\dfrac1x \).
3. So the derivative is \( \dfrac1x \) for all \( x\ne0 \).

Teacher's Note:
1. Remember \( \int\frac{dx}{x}=\log|x|+c \) as the reverse of this result.
2. Check the case \( x<0 \) separately.

Answer: (b) \( \log(ex) \)

Explanation:
1. \( y=x\log x \).
2. \( \dfrac{dy}{dx}=\log x+x\cdot\dfrac1x=\log x+1 \).
3. \( \log x+1=\log(ex) \).

Teacher's Note:
1. Bring down the exponent first: \( \log x^x=x\log x \).
2. Do not confuse this with the derivative of \( x^x \).

Answer: (c) \( \dfrac1{x\log x} \)

Explanation:
1. \( f[\log x]=\log(\log x) \).
2. Its derivative is \( \dfrac1{\log x}\cdot\dfrac1x \).
3. \( =\dfrac1{x\log x} \).

Teacher's Note:
1. Substitute the inner function into f first.
2. Then apply the chain rule.

Answer: (b) \( \dfrac1{\sqrt x(1-x)} \)

Explanation:
1. \( y=\log(1+\sqrt x)-\log(1-\sqrt x) \).
2. \( \dfrac{dy}{dx}=\dfrac1{2\sqrt x}\left[\dfrac1{1+\sqrt x}+\dfrac1{1-\sqrt x}\right]=\dfrac1{2\sqrt x}\cdot\dfrac2{1-x} \).
3. \( =\dfrac1{\sqrt x(1-x)} \).

Teacher's Note:
1. Split the logarithm of a quotient into a difference.
2. \( (1+\sqrt x)(1-\sqrt x)=1-x \).

Answer: (b) \( \csc x \)

Explanation:
1. \( \dfrac{1-\cos x}{1+\cos x}=\tan^2\dfrac x2 \), so the expression is \( \log\tan\dfrac x2 \).
2. Its derivative is \( \dfrac{\frac12\sec^2\frac x2}{\tan\frac x2}=\dfrac1{2\sin\frac x2\cos\frac x2} \).
3. \( =\dfrac1{\sin x}=\csc x \).

Teacher's Note:
1. Use the half-angle identities to simplify before differentiating.
2. The result is the standard \( \int\csc x\,dx \) in reverse.

Answer: (b) \( \sec x \)

Explanation:
1. \( \dfrac{\sec x\tan x+\sec^2x}{\sec x+\tan x} \).
2. Factor \( \sec x \) in the numerator: \( \dfrac{\sec x(\tan x+\sec x)}{\sec x+\tan x} \).
3. \( =\sec x \).

Teacher's Note:
1. This is the standard result behind \( \int\sec x\,dx \).
2. Factor the numerator to cancel with the denominator.

Answer: (c) \( \dfrac1{\sqrt{a^2+x^2}} \)

Explanation:
1. \( y=\log\left(x+\sqrt{a^2+x^2}\right)-\log a \).
2. \( \dfrac{dy}{dx}=\dfrac{1+\frac x{\sqrt{a^2+x^2}}}{x+\sqrt{a^2+x^2}}=\dfrac{\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}}}{x+\sqrt{a^2+x^2}} \).
3. \( =\dfrac1{\sqrt{a^2+x^2}} \).

Teacher's Note:
1. This is the standard derivative of \( \sinh^{-1}(x/a) \).
2. The constant \( \log a \) differentiates to zero.

Answer: (a) \( \dfrac1{\sin^{-1}x\sqrt{1-x^2}} \)

Explanation:
1. \( \dfrac d{dx}\log u=\dfrac{u'}u \) with \( u=\sin^{-1}x \).
2. \( u'=\dfrac1{\sqrt{1-x^2}} \).
3. So \( \dfrac{dy}{dx}=\dfrac1{\sin^{-1}x\sqrt{1-x^2}} \).

Teacher's Note:
1. Differentiate the inverse sine first and divide by \( \sin^{-1}x \).
2. No extra factor of x appears.

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

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

Teacher's Note:
1. Logarithmic differentiation handles a variable exponent.
2. The factor 2 comes from differentiating \( (\ln x)^2 \).

Answer: (c) \( 6x^5+6^x\log6 \)

Explanation:
1. \( \dfrac d{dx}x^6=6x^5 \) (power rule).
2. \( \dfrac d{dx}6^x=6^x\log6 \) (exponential rule).
3. Sum: \( 6x^5+6^x\log6 \).

Teacher's Note:
1. Do not apply the power rule to \( 6^x \): the variable is in the exponent.
2. \( \frac{d}{dx}a^x=a^x\log a \).

Answer: (a) \( e^x(\log\sin2x+2\cot2x) \)

Explanation:
1. Product rule: \( e^x\log\sin2x+e^x\cdot\dfrac{2\cos2x}{\sin2x} \).
2. \( =e^x(\log\sin2x+2\cot2x) \).

Teacher's Note:
1. The derivative of \( \log\sin2x \) is \( 2\cot2x \).
2. Factor out \( e^x \) at the end.

Answer: (a) y

Explanation:
1. The series is \( e^x \).
2. \( \dfrac{dy}{dx}=1+x+\dfrac{x^2}{2!}+\cdots=y \).
3. Differentiating term by term returns the same series.

Teacher's Note:
1. Recognise the exponential series.
2. Differentiating term by term reproduces the series.

Answer: (a) \( e^{x\sin x}(x\cos x+\sin x) \)

Explanation:
1. \( \dfrac d{dx}e^u=e^uu' \) with \( u=x\sin x \).
2. \( u'=\sin x+x\cos x \).
3. The result is \( e^{x\sin x}(x\cos x+\sin x) \).

Teacher's Note:
1. Use the product rule inside the exponent.
2. Write the factors in any order: the sum is the same.

Answer: (a) \( 2x^2e^{x^2}+e^{x^2} \)

Explanation:
1. Product rule: \( e^{x^2}+x\cdot e^{x^2}\cdot2x \).
2. \( =e^{x^2}(1+2x^2) \).
3. This equals \( 2x^2e^{x^2}+e^{x^2} \).

Teacher's Note:
1. The chain rule on \( e^{x^2} \) gives the factor \( 2x \).
2. Then multiply by the outside factor x.

Answer: (c) \( \dfrac{2\left(x^2+\log x\cdot x^{\log x}\right)}x \)

Explanation:
1. \( \dfrac d{dx}x^2=2x \). For \( z=x^{\log x} \): \( \log z=(\log x)^2 \), so \( z'=z\cdot\dfrac{2\log x}x \).
2. \( y'=2x+\dfrac{2\log x\cdot x^{\log x}}x \).
3. \( =\dfrac{2\left(x^2+\log x\cdot x^{\log x}\right)}x \).

Teacher's Note:
1. Differentiate the two terms separately, using logarithmic differentiation for the second.
2. Combine over the common denominator x.

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

Explanation:
1. Let \( u=e^x \). \( \ln y=\dfrac12[\ln(1+u)-\ln(1-u)] \).
2. \( \dfrac{y'}y=\dfrac12\,u\left[\dfrac1{1+u}+\dfrac1{1-u}\right]=\dfrac u{1-u^2} \).
3. \( y'=\dfrac u{(1-u)(1+u)}\sqrt{\dfrac{1+u}{1-u}}=\dfrac u{(1-u)\sqrt{1-u^2}} \).

Teacher's Note:
1. Use logarithmic differentiation for a square root of a quotient.
2. Combine \( \sqrt{1-u}\sqrt{1+u}=\sqrt{1-u^2} \).