JEE Mathematics Differentiation MCQs Set 05

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

Explanation:
1. \( \dfrac{x-x^{-1}}{x+x^{-1}}=\dfrac{x^2-1}{x^2+1}=-\dfrac{1-x^2}{1+x^2} \).
2. \( \cos^{-1}\left(-\dfrac{1-x^2}{1+x^2}\right)=\pi-2\tan^{-1}x \) for \( x>0 \).
3. Hence \( y'=-\dfrac{2}{1+x^2} \).

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) \( - \frac{x}{\sqrt{1 - x^{4}}} \)

Explanation:
1. \( y'=-\dfrac{1}{\sqrt{1-\frac{1+x^2}{2}}}\cdot\dfrac{x}{\sqrt2\sqrt{1+x^2}} \).
2. \( \sqrt{1-\frac{1+x^2}2}=\sqrt{\frac{1-x^2}{2}} \), so \( y'=-\dfrac{x}{\sqrt{1-x^2}\sqrt{1+x^2}}=-\dfrac{x}{\sqrt{1-x^4}} \).

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: (c) \( - \frac{1}{\sqrt{x} \left(2 x + 2\right)} \)

Explanation:
1. With \( \tan A=\sqrt a \), \( \tan B=\sqrt x \): \( y=\tan^{-1}\sqrt a-\tan^{-1}\sqrt x \).
2. \( y'=-\dfrac{1}{1+x}\cdot\dfrac{1}{2\sqrt x} \).

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) \( - \frac{x}{\sqrt{1 - x^{4}}} \)

Explanation:
1. Put \( x^2=\cos2\theta \): the argument becomes \( \dfrac{\sqrt2\cos\theta+\sqrt2\sin\theta}{\sqrt2\cos\theta-\sqrt2\sin\theta}=\tan\left(\dfrac\pi4+\theta\right) \).
2. So \( y=\dfrac\pi4+\dfrac12\cos^{-1}x^2 \), and \( y'=-\dfrac12\cdot\dfrac{2x}{\sqrt{1-x^4}}=-\dfrac{x}{\sqrt{1-x^4}} \).

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) \( \dfrac\pi4-1 \)

Explanation:
1. \( f'(x)=\tan^{-1}(e^{-2x})+(x+1)\cdot\dfrac{-2e^{-2x}}{1+e^{-4x}} \).
2. At \( x=0 \): \( \tan^{-1}1+1\cdot\dfrac{-2}{2}=\dfrac\pi4-1 \).

Teacher's Note:
1. Check each term of the product rule separately before combining them.
2. Keep the common exponential or trigonometric factor outside the bracket to compare with the options quickly.

Answer: (b) \( -1 \)

Explanation:
1. \( (1+x^{1/4})(1-x^{1/4})=1-x^{1/2} \).
2. Then \( y=(1-x^{1/2})(1+x^{1/2})=1-x \), so \( y'=-1 \).

Teacher's Note:
1. Simplify the inside expression where possible before applying the chain rule.
2. Work from the outside in and multiply by the derivative of the inner function at every layer.

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

Explanation:
1. Put \( x=\sin A \), \( y=\sin B \): \( \cos A+\cos B=a(\sin A-\sin B) \).
2. This gives \( \cot\dfrac{A-B}{2}=a \), so \( A-B=\text{constant} \).
3. Differentiating, \( \dfrac{dA}{dx}=\dfrac{dB}{dx} \), i.e. \( \dfrac{1}{\sqrt{1-x^2}}=\dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx} \), so \( \dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}} \).

Teacher's Note:
1. Remember the product rule on terms like xy and the chain rule on terms like sin y.
2. dy/dx equals minus the partial derivative in x divided by the partial derivative in y.

Answer: (d) \( \frac{1}{4 \sqrt{x} \sqrt{\sqrt{x} + 1}} \)

Explanation:
1. \( \dfrac{d}{dx}\sqrt{u}=\dfrac{u'}{2\sqrt u} \) with \( u=\sqrt x+1 \), \( u'=\dfrac1{2\sqrt x} \).
2. So the derivative is \( \dfrac{1}{4\sqrt x\sqrt{\sqrt x+1}} \).

Teacher's Note:
1. Simplify the inside expression where possible before applying the chain rule.
2. Work from the outside in and multiply by the derivative of the inner function at every layer.

Answer: (a) \( \frac{x \left(- \frac{1}{\sqrt{b^{2} + x^{2}}} + \frac{1}{\sqrt{a^{2} + x^{2}}}\right)}{a^{2} - b^{2}} \)

Explanation:
1. Rationalise: \( f(x)=\dfrac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{a^2-b^2} \).
2. Differentiate: \( f'(x)=\dfrac{1}{a^2-b^2}\left[\dfrac{x}{\sqrt{x^2+a^2}}-\dfrac{x}{\sqrt{x^2+b^2}}\right] \).

Teacher's Note:
1. Take logarithms first when the expression has many factors in numerator and denominator.
2. Simplify the quotient by algebra before differentiating if it reduces to a simple form.

Answer: (c) \( \frac{x^{2} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}}{y^{2}} \)

Explanation:
1. Put \( x^3=\sin A \), \( y^3=\sin B \): \( \cos A+\cos B=a^3(\sin A-\sin B) \), so \( \cot\dfrac{A-B}2=a^3 \) and \( A-B \) is constant.
2. Then \( \dfrac{dA}{dx}=\dfrac{dB}{dx} \): \( \dfrac{3x^2}{\sqrt{1-x^6}}=\dfrac{3y^2}{\sqrt{1-y^6}}\dfrac{dy}{dx} \).
3. So \( \dfrac{dy}{dx}=\dfrac{x^2}{y^2}\sqrt{\dfrac{1-y^6}{1-x^6}} \), option (c).

Teacher's Note:
1. dy/dx equals minus the partial derivative in x divided by the partial derivative in y.
2. Differentiate both sides with respect to x, treating y as a function of x, then collect the dy/dx terms.

Answer: (a) \( \frac{2 \sqrt{x} + 1}{4 \sqrt{x}} \)

Explanation:
1. \( y^2=x+\sqrt x \). Differentiate: \( 2y\dfrac{dy}{dx}=1+\dfrac1{2\sqrt x} \).
2. So \( y\dfrac{dy}{dx}=\dfrac12+\dfrac1{4\sqrt x}=\dfrac{2\sqrt x+1}{4\sqrt x} \).

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

Explanation:
1. Let \( u=\dfrac{2x-1}{x^2+1} \). Then \( \dfrac{dy}{dx}=f'(u)\dfrac{du}{dx}=\sin(u^2)\dfrac{du}{dx} \).
2. \( \dfrac{du}{dx}=\dfrac{2(x^2+1)-2x(2x-1)}{(x^2+1)^2}=\dfrac{-2x^2+2x+2}{(x^2+1)^2} \).
3. So \( \dfrac{dy}{dx}=\dfrac{-2x^2+2x+2}{(x^2+1)^2}\sin\left(\dfrac{2x-1}{x^2+1}\right)^2 \), where the square applies to the argument inside the sine.

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: (b) \( \dfrac{e^{ax}[a\sin(bx+c)-b\cos(bx+c)]}{\sin^2(bx+c)} \)

Explanation:
1. Quotient rule: \( \dfrac{ae^{ax}\sin(bx+c)-e^{ax}b\cos(bx+c)}{\sin^2(bx+c)} \).
2. This equals \( \dfrac{e^{ax}[a\sin(bx+c)-b\cos(bx+c)]}{\sin^2(bx+c)} \).

Teacher's Note:
1. Simplify the quotient by algebra before differentiating if it reduces to a simple form.
2. For a quotient, either use the quotient rule or rewrite it as a product with a negative power.

Answer: (a) \( \cos\left(\dfrac{5x+1}{10x^2-3}\right)\dfrac{d}{dx}\left(\dfrac{5x+1}{10x^2-3}\right) \)

Explanation:
1. Let \( u=\dfrac{5x+1}{10x^2-3} \). By the chain rule \( \dfrac{dy}{dx}=f'(u)\dfrac{du}{dx}=\cos u\cdot\dfrac{du}{dx} \).
2. This is option (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) \( -\dfrac{\sec^2(a^{1/x})\cdot(a^{1/x}\log_ea)}{x^2} \)

Explanation:
1. \( \dfrac{d}{dx}\tan u=\sec^2u\,u' \) with \( u=a^{1/x} \).
2. \( u'=a^{1/x}\log a\cdot\left(-\dfrac1{x^2}\right) \).
3. So the derivative is \( -\dfrac{\sec^2(a^{1/x})\,a^{1/x}\log a}{x^2} \).

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) \( \dfrac12\sqrt{\dfrac{1-\tan x}{1+\tan x}}\sec^2\left(\dfrac\pi4+x\right) \)

Explanation:
1. \( \dfrac{1+\tan x}{1-\tan x}=\tan\left(\dfrac\pi4+x\right) \), so \( y=\sqrt{\tan\left(\dfrac\pi4+x\right)} \).
2. \( y'=\dfrac{1}{2\sqrt{\tan(\frac\pi4+x)}}\sec^2\left(\dfrac\pi4+x\right) \) and \( \dfrac1{\sqrt{\tan(\frac\pi4+x)}}=\sqrt{\dfrac{1-\tan x}{1+\tan x}} \).
3. Hence option (a).

Teacher's Note:
1. Simplify the inside expression where possible before applying the chain rule.
2. Work from the outside in and multiply by the derivative of the inner function at every layer.

Answer: (a) \( \dfrac1{4\sqrt x}(\sec\sqrt x)^{3/2}\sin\sqrt x \)

Explanation:
1. \( y=(\sec\sqrt x)^{1/2} \). Then \( y'=\dfrac12(\sec\sqrt x)^{-1/2}\sec\sqrt x\tan\sqrt x\cdot\dfrac1{2\sqrt x} \).
2. \( \dfrac{\sec\sqrt x\tan\sqrt x}{(\sec\sqrt x)^{1/2}}=(\sec\sqrt x)^{1/2}\tan\sqrt x=(\sec\sqrt x)^{3/2}\sin\sqrt x \), giving option (a).

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) \( [f(m)+f''(m)]^2 \)

Explanation:
1. \( \dfrac{dx}{dm}=f'\cos m-f\sin m-f''\sin m-f'\cos m=-(f+f'')\sin m \).
2. \( \dfrac{dy}{dm}=f'\sin m+f\cos m+f''\cos m-f'\sin m=(f+f'')\cos m \).
3. Squaring and adding gives \( [f(m)+f''(m)]^2 \).

Teacher's Note:
1. For parametric equations, dy/dx equals (dy/dt) divided by (dx/dt).
2. Differentiate x and y separately with respect to the parameter and take the ratio.

Answer: (c) \( - \frac{4}{\log{\left(2 \right)}} \)

Explanation:
1. \( y=\dfrac{\log\tan x}{\log\sin x} \), so \( y'=\dfrac{\frac{\sec^2x}{\tan x}\log\sin x-\log\tan x\cdot\cot x}{(\log\sin x)^2} \).
2. At \( x=\dfrac\pi4 \): \( \log\tan x=0 \), \( \dfrac{\sec^2x}{\tan x}=2 \), \( \log\sin x=-\dfrac12\log2 \).
3. \( y'=\dfrac{2\cdot(-\frac12\log2)}{\frac14(\log2)^2}=-\dfrac{4}{\log2} \).

Teacher's Note:
1. Use y = e^(g log f) or logarithmic differentiation for functions of the form f(x)^g(x).
2. Differentiate the logarithm of y and multiply by y at the end.

Answer: (a) \( 0 \)

Explanation:
1. \( \dfrac{d}{dx}\log(\sqrt{x^2+1}-x)=\dfrac{\frac{x}{\sqrt{x^2+1}}-1}{\sqrt{x^2+1}-x}=-\dfrac1{\sqrt{x^2+1}} \).
2. Differentiate the left side: \( y'\sqrt{x^2+1}+\dfrac{xy}{\sqrt{x^2+1}}=-\dfrac{1}{\sqrt{x^2+1}} \).
3. Multiply by \( \sqrt{x^2+1} \): \( (x^2+1)y'+xy+1=0 \).

Teacher's Note:
1. Differentiate both sides with respect to x, treating y as a function of x, then collect the dy/dx terms.
2. Remember the product rule on terms like xy and the chain rule on terms like sin y.

Answer: (a) \( \frac{2 \sqrt{x} + e^{\sqrt{x}}}{2 \sqrt{x} \left(x + e^{\sqrt{x}}\right)} \)

Explanation:
1. \( y'=\dfrac{1+e^{\sqrt x}\cdot\frac1{2\sqrt x}}{x+e^{\sqrt x}} \).
2. Multiply numerator and denominator by \( 2\sqrt x \): \( \dfrac{2\sqrt x+e^{\sqrt x}}{2\sqrt x(x+e^{\sqrt x})} \).

Teacher's Note:
1. Derivatives of logarithmic expressions become easy once log laws split products, quotients and powers into sums.
2. Remember d/dx log u = u'/u and use log laws before differentiating.

Answer: (b) \( e^{\sqrt{1-x^2}}\left[\sec^2x-\dfrac{x\tan x}{\sqrt{1-x^2}}\right] \)

Explanation:
1. Product rule: \( e^{\sqrt{1-x^2}}\sec^2x+\tan x\cdot e^{\sqrt{1-x^2}}\cdot\dfrac{-x}{\sqrt{1-x^2}} \).
2. Factor out the exponential to get option (b).

Teacher's Note:
1. Check each term of the product rule separately before combining them.
2. Keep the common exponential or trigonometric factor outside the bracket to compare with the options quickly.

Answer: (b) \( \ln10\,(\tan x+x\sec^2x) \)

Explanation:
1. \( \dfrac{d}{dx}10^{x\tan x}=10^{x\tan x}\ln10\,(\tan x+x\sec^2x) \).
2. Multiplying by \( 10^{-x\tan x} \) leaves \( \ln10\,(\tan x+x\sec^2x) \).

Teacher's Note:
1. Differentiate the exponent first and then multiply by the original exponential.
2. Combine the exponential factor with the other factor using the product rule.

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

Explanation:
1. Put \( x=\cos2\theta \): \( \sqrt{\dfrac{1-x}{1+x}}=\tan\theta \), so \( 2\tan^{-1}\tan\theta=2\theta=\cos^{-1}x \).
2. Then \( y=\sin(\cos^{-1}x)=\sqrt{1-x^2} \) and \( y'=-\dfrac{x}{\sqrt{1-x^2}} \).

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: (b) \( e^{-1} \)

Explanation:
1. Put \( \log x=\tan\theta \): the function is \( \cos^{-1}\cos2\theta=2\theta=2\tan^{-1}(\log x) \) for \( \log x\ge0 \).
2. \( f'(x)=\dfrac{2}{1+(\log x)^2}\cdot\dfrac1x \); at \( x=e \): \( \dfrac{2}{2}\cdot\dfrac1e=\dfrac1e \).

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.