JEE Mathematics Differentiation MCQs Set 10

Answer: (a) 2

Explanation:
1. By the chain rule \( \frac{dy}{dx}=f'(x^2)\cdot2x \).
2. At x = 1: \( f'(1)\cdot2=\sqrt{2-1}\cdot2=2 \).

Teacher's Note:
1. Differentiate the composite function f(x^2).
2. Evaluate f' at x^2 = 1.

Answer: (b) \( \frac1e \)

Explanation:
1. Write \( f(x)=\frac{\ln(\ln x)}{\ln x} \).
2. Then \( f'(x)=\frac{\frac1x-\frac{\ln(\ln x)}{x}}{(\ln x)^2} \).
3. At x = e: \( \ln e=1,\ \ln(\ln e)=0 \), so \( f'(e)=\frac1e \).

Teacher's Note:
1. Change the base using ln.
2. Use the quotient rule.

Answer: (a) \( \frac2{19} \)

Explanation:
1. By the chain rule \( \frac{dy}{dt}=(3x^2-8)\frac{dx}{dt} \).
2. At x = 3: \( 3\cdot9-8=19 \), so \( \frac{dx}{dt}=\frac{2}{19} \).

Teacher's Note:
1. Relate dy/dt and dx/dt through dy/dx.
2. dy/dx = 3x^2 - 8.

Answer: (c) 1

Explanation:
1. Differentiate: \( \frac{dx}{dt}=-\frac{2}{t^3}-\frac{3}{t^4}=-\frac{2t+3}{t^4} \) and \( \frac{dy}{dt}=-\frac3{t^3}-\frac2{t^2}=-\frac{2t+3}{t^3} \).
2. So \( \frac{dy}{dx}=t \).
3. Then \( x\,t^3-t=\frac{(1+t)}{t^3}t^3-t=1 \).

Teacher's Note:
1. Find dy/dx by the parametric formula.
2. The common factor (2t + 3) cancels.

Answer: (c) \( -\frac1{2at^3} \)

Explanation:
1. First derivative: \( \frac{dy}{dx}=\frac{2a}{2at}=\frac1t \).
2. Second derivative: \( \frac{d}{dt}\left(\frac1t\right)\Big/\frac{dx}{dt}=-\frac{1}{t^2}\cdot\frac{1}{2at}=-\frac1{2at^3} \).

Teacher's Note:
1. Differentiate dy/dx with respect to t, then divide by dx/dt.
2. Do not forget the extra factor 1/(2at).

Answer: (d) \( -\frac1{(1+x)^2} \)

Explanation:
1. Isolate: \( x\sqrt{1+y}=-y\sqrt{1+x} \). Squaring gives \( x^2(1+y)=y^2(1+x) \), i.e. \( x^2-y^2+xy(x-y)=0 \).
2. Factor: \( (x-y)(x+y+xy)=0 \). For \( x\ne y \) this gives \( y=-\frac{x}{1+x} \).
3. Then \( \frac{dy}{dx}=-\frac{1}{(1+x)^2} \).

Teacher's Note:
1. Square both sides and factor.
2. The factor x - y = 0 would not satisfy the original equation for x not 0.

Answer: (b) \( \frac{1+a^2}{a^{10}} \)

Explanation:
1. Since g is the inverse of f and \( g(2)=a \), we have \( f(a)=2 \).
2. Then \( g'(2)=\frac1{f'(g(2))}=\frac1{f'(a)}=\frac{1+a^2}{a^{10}} \).

Teacher's Note:
1. Use g'(y) = 1 / f'(g(y)).
2. g(2) = a means f(a) = 2.

Answer: (d) \( \frac yx \)

Explanation:
1. The equation is homogeneous of degree 2, so it factorises into two lines through the origin: \( y=mx \) with constant m.
2. On each line \( \frac{dy}{dx}=m=\frac yx \).

Teacher's Note:
1. A homogeneous equation of degree 2 represents lines through the origin.
2. The slope of such a line is y/x.

Answer: (d) 2 and -1

Explanation:
1. Since \( 1+x^2+x^4=(1+x+x^2)(1-x+x^2) \), the fraction equals \( x^2-x+1 \).
2. Its derivative is \( 2x-1 \), so a = 2 and b = -1.

Teacher's Note:
1. Factorise 1 + x^2 + x^4.
2. Cancel the common factor before differentiating.

Answer: (b) \( e^{x+e^x} \)

Explanation:
1. From \( e^{f(x)}=\ln x \) we get \( f(x)=\ln(\ln x) \).
2. The inverse satisfies \( y=\ln(\ln g) \), so \( g(y)=e^{e^y} \).
3. Then \( g'(x)=e^{e^x}\cdot e^x=e^{x+e^x} \).

Teacher's Note:
1. Solve for f(x) first, then invert.
2. Differentiate e^(e^x).

Answer: (a) \( \frac yx \)

Explanation:
1. Take logarithms: \( m\ln x+n\ln y=(m+n)\ln(x+y) \).
2. Differentiate: \( \frac mx+\frac ny\,y'=\frac{(m+n)(1+y')}{x+y} \).
3. Collecting terms, \( y'\left[\frac ny-\frac{m+n}{x+y}\right]=\frac{m+n}{x+y}-\frac mx \), which simplifies to \( y'\cdot\frac{nx-my}{y(x+y)}=\frac{nx-my}{x(x+y)} \), so \( y'=\frac yx \).

Teacher's Note:
1. Logarithmic differentiation.
2. Both brackets contain the factor (nx - my).

Answer: (b) \( -\frac1{14} \)

Explanation:
1. Replace x by 1/x: \( 8f(1/x)+6f(x)=\frac1x+5 \).
2. Solving the pair: \( f(x)=\frac{4x^2+5x-3}{14x} \).
3. Then \( y=x^2f(x)=\frac{x(4x^2+5x-3)}{14}=\frac{4x^3+5x^2-3x}{14} \) and \( \frac{dy}{dx}=\frac{12x^2+10x-3}{14} \), which is \( \frac{12-10-3}{14}=-\frac1{14} \) at x = -1.

Teacher's Note:
1. Eliminate f(1/x) between the two equations.
2. Differentiate the resulting polynomial.

Answer: (b) \( \frac1{\sqrt2} \)

Explanation:
1. Since \( \sec(\tan^{-1}x)=\sqrt{1+x^2} \), \( y'=\frac{x}{\sqrt{1+x^2}} \).
2. At x = 1 it is \( \frac1{\sqrt2} \).

Teacher's Note:
1. Draw a right triangle with tan theta = x.
2. Then sec theta = sqrt(1 + x^2).

Answer: (d) \( f''(e^x)e^{2x}+f'(e^x)e^x \)

Explanation:
1. First derivative: \( f'(e^x)e^x \).
2. Second derivative by the product rule: \( f''(e^x)e^x\cdot e^x+f'(e^x)e^x=f''(e^x)e^{2x}+f'(e^x)e^x \).

Teacher's Note:
1. Apply the chain rule twice.
2. The factor e^x also has to be differentiated.

Answer: (b) 0

Explanation:
1. Differentiating a determinant row by row, the first two rows give determinants with two equal rows (zero).
2. The third row becomes \( (f''',g''',h''') \), which is zero for polynomials of degree 2.
3. So \( F'(x)=0 \).

Teacher's Note:
1. Differentiate one row at a time.
2. The third derivative of a quadratic is 0.

Answer: (b) Statement 1 is false and Statement 2 is true

Explanation:
1. With \( y=x^2 \): \( \frac{d^2y}{dx^2}=2 \). Also \( x=\sqrt y \) gives \( \frac{d^2x}{dy^2}=-\frac1{4y^{3/2}}=-\frac1{4x^3} \).
2. The product is \( -\frac1{2x^3} \), not 1, so Statement 1 is false.
3. Statement 2 is the standard formula, which is true.

Teacher's Note:
1. Compute both second derivatives explicitly.
2. Check the product with the formula in Statement 2.

Answer: (c) -1

Explanation:
1. At x = 1: \( 1-2\cot y-1=0 \), so \( y=\frac\pi2 \).
2. Differentiate implicitly: \( x^{2x}(2\ln x+2)-2x^x(\ln x+1)\cot y+2x^x\csc^2y\,y'=0 \).
3. At (1, pi/2): \( 1\cdot2-0+2\cdot1\cdot1\cdot y'=0 \), so \( y'=-1 \).

Teacher's Note:
1. Find y at x = 1 first.
2. d/dx x^x = x^x (ln x + 1).

Answer: (d) -4

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

Teacher's Note:
1. Apply the chain rule from the outside in.
2. 2f(0) + 2 = 0.

Answer: (b) \( -\frac{d^2y}{dx^2}\left(\frac{dy}{dx}\right)^{-3} \)

Explanation:
1. We have \( \frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1} \).
2. Differentiate with respect to y: \( \frac{d^2x}{dy^2}=-\left(\frac{dy}{dx}\right)^{-2}\frac{d^2y}{dx^2}\cdot\frac{dx}{dy} \).
3. Substituting dx/dy gives \( -\frac{d^2y}{dx^2}\left(\frac{dy}{dx}\right)^{-3} \).

Teacher's Note:
1. Differentiate (dy/dx)^(-1) with respect to y using the chain rule.
2. d/dy = (dx/dy) d/dx.