JEE Mathematics Methods of Differentiation MCQs Set 07

Answer: (c) \( e \)

Explanation:
1. At \( (1,\pi) \): \( \sin\pi+e^{1}=e \), so the point lies on the curve.
2. Differentiate: \( \cos y\,y'+e^{-x\cos y}(-\cos y+x\sin y\,y')=0 \).
3. At the point: \( -y'+e(1+0)=0 \), so \( y'=e \).

Teacher's Note:
1. Check that the point lies on the curve first.
2. Then substitute cos pi = -1 and sin pi = 0.

Answer: (b) \( \frac{2\sin x^2}{3x\cos x^3} \)

Explanation:
1. \( \frac{du}{dx}=f'(x^2)\cdot2x=2x\sin x^2 \).
2. \( \frac{dv}{dx}=g'(x^3)\cdot3x^2=3x^2\cos x^3 \).
3. Ratio: \( \frac{2\sin x^2}{3x\cos x^3} \).

Teacher's Note:
1. Differentiate both with respect to x and divide.
2. Evaluate f' and g' at the inner functions.

Answer: (a) 16

Explanation:
1. Chain rule: \( y'(0)=f'(f^3(0))\,f'(f^2(0))\,f'(f(0))\,f'(0) \).
2. Since \( f(0)=0 \), every inner value is \( 0 \), so \( y'(0)=f'(0)^4=2^4=16 \).

Teacher's Note:
1. Because 0 is a fixed point, every layer is evaluated at 0.
2. The derivative is the fourth power of f'(0).

Answer: (d) 10

Explanation:
1. \( u=(s+3t)^2+(2s-t)^2=s^2+6st+9t^2+4s^2-4st+t^2=5s^2+2st+10t^2 \).
2. \( \frac{du}{ds}=10s+2t \) and \( \frac{d^2u}{ds^2}=10 \).

Teacher's Note:
1. Substitute first, then differentiate with t held constant.
2. Only the s squared coefficient matters for the second derivative.

Answer: (c) \( \frac{5}{16t^6} \)

Explanation:
1. \( \frac{dy}{dx}=\frac{10t^9}{8t^7}=\frac54t^2 \).
2. \( \frac{d^2y}{dx^2}=\frac{\frac52t}{8t^7}=\frac{5}{16t^6} \).

Teacher's Note:
1. Divide the derivative of dy/dx with respect to t by dx/dt.
2. Simplify the powers of t.

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

Explanation:
1. \( \sin(\cot^{-1}x)=\frac{1}{\sqrt{1+x^2}}=u \).
2. \( \cos^2(\tan^{-1}u)=\frac{1}{1+u^2}=\frac{1+x^2}{2+x^2} \).
3. Differentiate: \( \frac{2x(2+x^2)-2x(1+x^2)}{(2+x^2)^2}=\frac{2x}{(x^2+2)^2} \).

Teacher's Note:
1. Simplify from the inside using right-triangle ratios.
2. The final quotient simplifies nicely.

Answer: (d) \( -1 \)

Explanation:
1. Let \( u=\frac{x^x-x^{-x}}{2} \); \( u(1)=0 \).
2. \( u'=\frac{x^x(1+\log x)+x^{-x}(1+\log x)}{2} \), so \( u'(1)=1 \).
3. \( f'(1)=\frac{-u'}{1+u^2}=-1 \).

Teacher's Note:
1. Evaluate u and u' at x = 1 separately.
2. The derivative of cot inverse is minus one over (1 + u^2).

Answer: (d) \( \frac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x} \)

Explanation:
1. The pattern repeats every two levels: \( y=\frac{\sin x}{1+\frac{\cos x}{1+y}}=\frac{(1+y)\sin x}{1+y+\cos x} \).
2. Clear fractions: \( y^2+y(1+\cos x-\sin x)-\sin x=0 \).
3. Differentiate: \( y'(2y+1+\cos x-\sin x)=(1+y)\cos x+y\sin x \).

Teacher's Note:
1. Find the repeating unit and call it y.
2. Then differentiate the resulting quadratic relation implicitly.

Answer: (a) \( \frac{\mu-17v}{17\mu-v} \)

Explanation:
1. Square: \( \frac v\mu+\frac\mu v+2=36 \), so \( v^2+\mu^2=34\mu v \).
2. Differentiate with respect to \( \mu \): \( 2vv'+2\mu=34v+34\mu v' \).
3. \( v'(2v-34\mu)=34v-2\mu \), so \( v'=\frac{17v-\mu}{v-17\mu}=\frac{\mu-17v}{17\mu-v} \).

Teacher's Note:
1. Squaring converts the root equation into a homogeneous quadratic.
2. Multiply numerator and denominator by -1 to match the form.

Answer: (c) 0

Explanation:
1. Let \( L=\log x \). The first term is \( \tan^{-1}\frac{1-2L}{1+2L}=\frac\pi4-\tan^{-1}2L \).
2. The second is \( \tan^{-1}3+\tan^{-1}2L \) since \( \frac{3+2L}{1-6L}=\frac{3+2L}{1-3\cdot2L} \).
3. The sum is the constant \( \frac\pi4+\tan^{-1}3 \), so \( y''=0 \).

Teacher's Note:
1. Recognise the tan addition and subtraction patterns.
2. The log terms cancel, leaving a constant.

Answer: (a) \( \frac12 \)

Explanation:
1. With \( \sqrt{1\pm\sin x}=\left|\cos\frac x2\pm\sin\frac x2\right| \), the bracket equals \( \frac{2\cos\frac x2}{2\sin\frac x2}=\cot\frac x2 \).
2. So \( y=\frac x2 \) and \( y'=\frac12 \).

Teacher's Note:
1. Write 1 plus or minus sin x as a perfect square of half-angle terms.
2. The range 0 < x < pi/2 settles the signs.

Answer: (d) 2

Explanation:
1. \( y'=-\frac xy \).
2. \( y''=-\frac{y-xy'}{y^2}=-\frac{y+x^2/y}{y^2}=-\frac{x^2+y^2}{y^3}=-\frac{2}{y^3} \).
3. Comparing with \( y''=-Ay^{-3} \) gives \( A=2 \).

Teacher's Note:
1. Combine the terms over y cubed and use the given circle equation.
2. The result is a neat power of y.

Answer: (a) \( e\,e^{x^e}x^{e-1}+x^{e^x}e^x\left(\frac1x+\log x\right)+e^{x^x}x^x(1+\log x) \)

Explanation:
1. First term: \( e^{x^e}\cdot ex^{e-1} \).
2. Second: \( x^{e^x}\left[e^x\log x+\frac{e^x}{x}\right] \).
3. Third: \( e^{x^x}\cdot x^x(1+\log x) \).
4. Add the three.

Teacher's Note:
1. Each term needs its own chain or logarithmic differentiation.
2. The sign inside the brackets is plus in every case.

Answer: (c) \( n\cot n\theta \)

Explanation:
1. Take the logarithm of both sides and differentiate with respect to \( \theta \).
2. Left side: the sum of the cotangents. Right side: \( n\cot n\theta \).
3. Hence the sum is \( n\cot n\theta \).

Teacher's Note:
1. Logarithmic differentiation of a product gives a sum of derivative ratios.
2. The constant 2^(n-1) disappears.

Answer: (d) 2

Explanation:
1. \( y=\sqrt{\cot^2\theta}=|\cot\theta| \).
2. At \( \theta=\frac{3\pi}4 \), \( \cot\theta=-1<0 \), so \( y=-\cot\theta \) near this point.
3. \( y'=\mathrm{cosec}^2\theta=2 \).

Teacher's Note:
1. Remember the absolute value from the square root.
2. Fix its sign at the given angle before differentiating.

Answer: (c) \( \frac{\sqrt3-1}{2} \)

Explanation:
1. At \( x=\frac{2\pi}3 \), \( \cos x<0 \) and \( \sin x>0 \), so \( y=-\cos x+\sin x \) nearby.
2. \( y'=\sin x+\cos x=\frac{\sqrt3}{2}-\frac12 \).

Teacher's Note:
1. Remove each modulus using the sign at the point.
2. Differentiate the resulting expression.

Answer: (c) 2

Explanation:
1. \( \frac{dy}{dx}=f'(x^2)\cdot2x=2x\sqrt{2x^4-1} \).
2. At \( x=1 \): \( 2\sqrt{1}=2 \).

Teacher's Note:
1. Evaluate f' at x squared.
2. Then multiply by 2x.

Answer: (d) \( \frac34 \)

Explanation:
1. With \( u=\sqrt{\frac{1+x}{2}} \), \( \cos^{-1}(\sin u)=\frac\pi2-u \) for \( u\in[0,\frac\pi2] \), so its derivative is \( -u'=-\frac{1}{4u} \).
2. At \( x=1 \), \( u=1 \), giving \( -\frac14 \).
3. \( \frac{d}{dx}x^x=x^x(1+\log x)=1 \) at \( x=1 \).
4. Total: \( 1-\frac14=\frac34 \).

Teacher's Note:
1. Use cos inverse of sin u equals pi/2 minus u on the principal range.
2. Add the two separate derivatives.

Answer: (b) \( -1 \)

Explanation:
1. For \( x\in[\pi,2\pi] \), \( \cos^{-1}(\cos x)=2\pi-x \).
2. So \( y'=-1 \).

Teacher's Note:
1. The principal value of cos inverse lies in [0, pi].
2. Use the reflected branch for angles in the third and fourth quadrants.

Answer: (b) \( -2\tan\left(\frac{3x+4}{5x+6}\right)^2\cdot\frac{1}{(5x+6)^2} \)

Explanation:
1. Chain rule: \( y'=f'(u)\cdot u' \) with \( u=\frac{3x+4}{5x+6} \).
2. \( u'=\frac{3(5x+6)-5(3x+4)}{(5x+6)^2}=\frac{-2}{(5x+6)^2} \).
3. So \( y'=-2\tan(u^2)\cdot\frac{1}{(5x+6)^2} \).

Teacher's Note:
1. f' must be evaluated at u, so its argument becomes u squared.
2. The determinant 3*6 - 5*4 = -2 gives the sign.

Answer: (b) \( 4n^2 \)

Explanation:
1. \( \frac{dx}{d\theta}=-\cot\theta(\mathrm{cosec}\theta+\sin\theta) \) and \( \frac{dy}{d\theta}=-n\cot\theta(\mathrm{cosec}^n\theta+\sin^n\theta) \).
2. Also \( x^2+4=(\mathrm{cosec}\theta+\sin\theta)^2 \) and \( y^2+4=(\mathrm{cosec}^n\theta+\sin^n\theta)^2 \).
3. So \( (x^2+4)\left(\frac{dy}{dx}\right)^2=n^2(y^2+4) \), and subtracting \( n^2y^2 \) leaves \( 4n^2 \).

Teacher's Note:
1. Since cosec times sin equals 1, (a + b)^2 = (a - b)^2 + 4.
2. This produces the constant 4 in each bracket.

Answer: (d) \( 6x^2 \)

Explanation:
1. Expand along the first column: \( x(12x^2-6x^2)-1\cdot(6x^3-2x^3)=6x^3-4x^3=2x^3 \).
2. \( f'(x)=6x^2 \).

Teacher's Note:
1. Expand the determinant first when the entries are simple polynomials.
2. Differentiate the resulting expression.

Answer: (c) \( -1 \)

Explanation:
1. At \( x=1 \): \( 1-2\cot y-1=0 \), so \( \cot y=0 \) and \( y=\frac\pi2 \).
2. Differentiate: \( x^{2x}(2\log x+2)-2\left[x^x(1+\log x)\cot y-x^x\mathrm{cosec}^2y\,y'\right]=0 \).
3. At \( (1,\frac\pi2) \): \( 2-2[0-y']=0 \), so \( y'=-1 \).

Teacher's Note:
1. Find y at x = 1 first.
2. Both power terms are differentiated logarithmically.

Answer: (b) \( -4 \)

Explanation:
1. \( g'(x)=2f(2f(x)+2)\cdot f'(2f(x)+2)\cdot2f'(x) \).
2. At \( x=0 \): \( 2f(0)+2=0 \), so \( g'(0)=2f(0)f'(0)\cdot2f'(0)=2(-1)(1)(2)(1)=-4 \).

Teacher's Note:
1. Evaluate the inner expression 2f(0) + 2 first; it equals 0.
2. Then every factor is known from the data.