JEE Mathematics Differentiation MCQs Set 06

Answer: (a) \( \dfrac1\theta \)

Explanation:
1. Let \( u=x-\alpha \) and \( \phi=\cos^{-1}\dfrac{a\cos u+b}{a+b\cos u} \). Then \( \cos\phi=\dfrac{a\cos u+b}{a+b\cos u} \).
2. \( \tan^2\dfrac\phi2=\dfrac{1-\cos\phi}{1+\cos\phi}=\dfrac{(a-b)(1-\cos u)}{(a+b)(1+\cos u)}=\dfrac{a-b}{a+b}\tan^2\dfrac u2 \).
3. So \( \tan\dfrac\phi2=\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac u2 \), and differentiating gives \( \dfrac{d\phi}{dx}=\dfrac{\sqrt{a^2-b^2}}{a+b\cos u} \).
4. Therefore \( \dfrac{dy}{dx}=\dfrac1{\sqrt{a^2-b^2}}\cdot\dfrac{\sqrt{a^2-b^2}}{\theta}=\dfrac1\theta \).

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. \( \sec x+\tan x=\tan\left(\dfrac\pi4+\dfrac x2\right) \).
2. So the function is \( \dfrac\pi4+\dfrac x2 \) and the derivative is \( \dfrac12 \).

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) \( \frac{1}{2} \)

Explanation:
1. \( \sqrt{\dfrac{1-\cos x}{1+\cos x}}=\tan\dfrac x2 \), so the function is \( \dfrac x2 \) on \( 0<x<\pi \).
2. The 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) \( \frac{1}{2 \sqrt{1 - x^{2}}} \)

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

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

Explanation:
1. \( \cot^{-1}\sqrt{\dfrac{1-x}{1+x}} \): let \( x=\cos2\theta \), then \( \sqrt{\dfrac{1-x}{1+x}}=\tan\theta \) and \( \cot^{-1}\tan\theta=\dfrac\pi2-\theta \).
2. \( \sin^2\left(\dfrac\pi2-\theta\right)=\cos^2\theta=\dfrac{1+x}{2} \).
3. 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) \( \frac{3 a}{a^{2} + x^{2}} \)

Explanation:
1. Put \( x=a\tan\theta \): the argument is \( \dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\tan3\theta \).
2. So \( y=3\theta=3\tan^{-1}\dfrac xa \) and \( y'=\dfrac{3}{1+x^2/a^2}\cdot\dfrac1a=\dfrac{3a}{a^2+x^2} \).

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) \( \sec{\left(x \right)} \)

Explanation:
1. \( \dfrac{d}{dx}\sinh^{-1}u=\dfrac{u'}{\sqrt{1+u^2}} \) with \( u=\tan x \).
2. \( y'=\dfrac{\sec^2x}{\sqrt{1+\tan^2x}}=\dfrac{\sec^2x}{\sec x}=\sec x \) for \( \cos x>0 \).

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: (b) \( \frac{y + \sin{\left(2 x \right)}}{- x + 2 \sin{\left(y \right)}} \)

Explanation:
1. Differentiate: \( 2\sin x\cos x-2\sin y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0 \).
2. \( (x-2\sin y)\dfrac{dy}{dx}=-(y+\sin2x) \), so \( \dfrac{dy}{dx}=\dfrac{y+\sin2x}{2\sin y-x} \).

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) \( \frac{y \left(2 x y - y^{2} \cos{\left(x y \right)} - 1\right)}{x y^{2} \cos{\left(x y \right)} - x + y^{2}} \)

Explanation:
1. Differentiate: \( \cos(xy)\left(y+x\dfrac{dy}{dx}\right)+\dfrac{y-x\frac{dy}{dx}}{y^2}=2x-\dfrac{dy}{dx} \).
2. Multiply by \( y^2 \) and collect \( \dfrac{dy}{dx} \): \( \dfrac{dy}{dx}\left[xy^2\cos(xy)-x+y^2\right]=2xy^2-y^3\cos(xy)-y \).
3. So \( \dfrac{dy}{dx}=\dfrac{y[2xy-y^2\cos(xy)-1]}{xy^2\cos(xy)+y^2-x} \).

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: (c) \( \frac{\sin^{2}{\left(a + y \right)}}{\sin{\left(a \right)}} \)

Explanation:
1. \( x=\dfrac{\sin y}{\sin(a+y)} \), so \( \dfrac{dx}{dy}=\dfrac{\cos y\sin(a+y)-\sin y\cos(a+y)}{\sin^2(a+y)}=\dfrac{\sin a}{\sin^2(a+y)} \).
2. Therefore \( \dfrac{dy}{dx}=\dfrac{\sin^2(a+y)}{\sin a} \).

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: (a) \( x^{x} \log{\left(e x \right)} \)

Explanation:
1. \( \log y=x\log x \), so \( \dfrac{y'}{y}=\log x+1=\log(ex) \).
2. Hence \( y'=x^x\log(ex) \).

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) \( \dfrac y2\left[\dfrac1{x-a}+\dfrac1{x-b}-\dfrac1{x-c}-\dfrac1{x-d}\right] \)

Explanation:
1. \( \log y=\dfrac12[\log(x-a)+\log(x-b)-\log(x-c)-\log(x-d)] \).
2. Differentiate: \( \dfrac{y'}{y}=\dfrac12\left[\dfrac1{x-a}+\dfrac1{x-b}-\dfrac1{x-c}-\dfrac1{x-d}\right] \).

Teacher's Note:
1. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.
2. Use y = e^(g log f) or logarithmic differentiation for functions of the form f(x)^g(x).

Answer: (a) \( \dfrac{x\cos x\log x+\sin x}{x}\cdot x^{\sin x} \)

Explanation:
1. \( \log y=\sin x\log x \), so \( \dfrac{y'}{y}=\cos x\log x+\dfrac{\sin x}{x} \).
2. \( y'=y\cdot\dfrac{x\cos x\log x+\sin x}{x}=\dfrac{x\cos x\log x+\sin x}{x}\cdot x^{\sin x} \).

Teacher's Note:
1. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.
2. Use y = e^(g log f) or logarithmic differentiation for functions of the form f(x)^g(x).

Answer: (b) \( -1 \)

Explanation:
1. Differentiate: \( 2^x\ln2+2^y\ln2\dfrac{dy}{dx}=2^{x+y}\ln2\left(1+\dfrac{dy}{dx}\right) \).
2. At \( x=y=1 \): \( 2+2y'=4(1+y') \), so \( -2y'=2 \) and \( y'=-1 \).

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) \( \log x\,[\log(ex)]^{-2} \)

Explanation:
1. Take logs: \( y\log x=x-y \), so \( y=\dfrac{x}{1+\log x} \).
2. \( \dfrac{dy}{dx}=\dfrac{(1+\log x)-1}{(1+\log x)^2}=\dfrac{\log x}{[\log(ex)]^2} \).

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: (c) \( 2 \)

Explanation:
1. Let \( u=(\tan x)^{\tan x} \) so \( y=(\tan x)^u \). At \( x=\dfrac\pi4 \): \( \tan x=1 \), \( u=1 \), \( y=1 \).
2. \( \log y=u\log\tan x \), so \( \dfrac{y'}{y}=u'\log\tan x+u\cdot\dfrac{\sec^2x}{\tan x} \).
3. At \( \dfrac\pi4 \): \( \log\tan x=0 \), so \( \dfrac{y'}{y}=1\cdot\dfrac{2}{1}=2 \) and \( y'=2 \).

Teacher's Note:
1. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.
2. Use y = e^(g log f) or logarithmic differentiation for functions of the form f(x)^g(x).

Answer: (a) \( \frac{y}{x} \)

Explanation:
1. Take logs: \( p\log x+q\log y=(p+q)\log(x+y) \).
2. Differentiate: \( \dfrac px+\dfrac qy\dfrac{dy}{dx}=\dfrac{p+q}{x+y}\left(1+\dfrac{dy}{dx}\right) \).
3. \( \dfrac{dy}{dx}\left(\dfrac qy-\dfrac{p+q}{x+y}\right)=\dfrac{p+q}{x+y}-\dfrac px \) gives \( \dfrac{dy}{dx}\cdot\dfrac{qx-py}{y(x+y)}=\dfrac{qx-py}{x(x+y)} \), so \( \dfrac{dy}{dx}=\dfrac yx \).

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) \( y\csc^2x\,(1-\log\tan x) \)

Explanation:
1. \( \log y=\cot x\log\tan x \), so \( \dfrac{y'}{y}=-\csc^2x\log\tan x+\cot x\cdot\dfrac{\sec^2x}{\tan x} \).
2. \( \cot x\cdot\dfrac{\sec^2x}{\tan x}=\dfrac{\sec^2x}{\tan^2x}=\csc^2x \).
3. So \( y'=y\csc^2x\,(1-\log\tan x) \).

Teacher's Note:
1. Differentiate the logarithm of y and multiply by y at the end.
2. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.

Answer: (d) \( \dfrac{e^x[1+(x-2)\log x]}{x^3} \)

Explanation:
1. Quotient rule: \( y'=\dfrac{\left(e^x\log x+\frac{e^x}{x}\right)x^2-2xe^x\log x}{x^4} \).
2. \( =\dfrac{e^x[x\log x+1-2\log x]}{x^3}=\dfrac{e^x[1+(x-2)\log x]}{x^3} \).

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: (a) \( y\left[\dfrac1{2x}+\dfrac4{2x+3}-\dfrac1{2(x+1)}\right] \)

Explanation:
1. \( \log y=\dfrac12\log x+2\log(2x+3)-\dfrac12\log(x+1) \).
2. \( \dfrac{y'}{y}=\dfrac1{2x}+\dfrac4{2x+3}-\dfrac1{2(x+1)} \).

Teacher's Note:
1. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.
2. Use y = e^(g log f) or logarithmic differentiation for functions of the form f(x)^g(x).

Answer: (b) \( \dfrac{2(x-\sin x)^{3/2}}{\sqrt x}\left[\dfrac32\cdot\dfrac{1-\cos x}{x-\sin x}-\dfrac1{2x}\right] \)

Explanation:
1. \( \log y=\log2+\dfrac32\log(x-\sin x)-\dfrac12\log x \).
2. \( \dfrac{y'}{y}=\dfrac32\cdot\dfrac{1-\cos x}{x-\sin x}-\dfrac1{2x} \), so option (b).

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: (b) \( (x-2)^{x-1}[(x-2)\log(x-2)+x] \)

Explanation:
1. \( \log y=x\log(x-2) \), so \( \dfrac{y'}{y}=\log(x-2)+\dfrac{x}{x-2} \).
2. \( y'=(x-2)^x\left[\log(x-2)+\dfrac x{x-2}\right]=(x-2)^{x-1}[(x-2)\log(x-2)+x] \).

Teacher's Note:
1. Differentiate the logarithm of y and multiply by y at the end.
2. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.

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

Explanation:
1. \( \log y=a^x\log x \), so \( \dfrac{y'}{y}=a^x\ln a\log x+\dfrac{a^x}x \).
2. \( y'=x^{a^x}\left[\dfrac{a^x}x+a^x\log a\log x\right] \).

Teacher's Note:
1. For a variable base and variable exponent take logarithms of both sides and differentiate implicitly.
2. Use y = e^(g log f) or logarithmic differentiation for functions of the form f(x)^g(x).

Answer: (a) \( \frac{b \tan{\left(t \right)}}{a} \)

Explanation:
1. \( x=a\sin2\theta+\dfrac a2\sin4\theta \), so \( \dfrac{dx}{d\theta}=2a\cos2\theta+2a\cos4\theta=4a\cos3\theta\cos\theta \).
2. \( y=b\cos2\theta-\dfrac b2(1+\cos4\theta) \), so \( \dfrac{dy}{d\theta}=-2b\sin2\theta+2b\sin4\theta=4b\cos3\theta\sin\theta \).
3. Hence \( \dfrac{dy}{dx}=\dfrac{b\tan\theta}{a} \).

Teacher's Note:
1. Simplify the ratio with trigonometric identities before substituting the given parameter value.
2. For parametric equations, dy/dx equals (dy/dt) divided by (dx/dt).

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

Explanation:
1. \( \dfrac{dx}{dt}=\dfrac{2(1-t^2)}{(1+t^2)^2} \) and \( \dfrac{dy}{dt}=\dfrac{-4t}{(1+t^2)^2} \).
2. \( \dfrac{dy}{dx}=\dfrac{-4t}{2(1-t^2)}=\dfrac{-2t}{1-t^2}=\dfrac{2t}{t^2-1} \).

Teacher's Note:
1. Simplify the ratio with trigonometric identities before substituting the given parameter value.
2. For parametric equations, dy/dx equals (dy/dt) divided by (dx/dt).