JEE Mathematics Differentiation MCQs Set 07

Answer: (a) \( -1 \)

Explanation:
1. \( \dfrac{dx}{d\theta}=-4a\cos^3\theta\sin\theta \) and \( \dfrac{dy}{d\theta}=4a\sin^3\theta\cos\theta \).
2. \( \dfrac{dy}{dx}=-\tan^2\theta \). At \( \theta=\dfrac{3\pi}4 \): \( \tan\theta=-1 \), so \( \dfrac{dy}{dx}=-1 \).

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).

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

Explanation:
1. \( \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac1t \).
2. At \( t=2 \), \( \dfrac{dy}{dx}=\dfrac12 \).

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: (a) \( -\pi/6 \)

Explanation:
1. \( \dfrac{dy}{dt}=\dfrac\pi2\cos\dfrac\pi2t-\dfrac\pi2\sin\dfrac\pi2t \) and \( \dfrac{dx}{dt}=2t+1 \).
2. At \( t=1 \): \( \dfrac{dy}{dt}=0-\dfrac\pi2=-\dfrac\pi2 \), \( \dfrac{dx}{dt}=3 \).
3. So \( \dfrac{dy}{dx}=-\dfrac\pi6 \).

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: (a) \( \frac{y}{1 - y} \)

Explanation:
1. The pattern repeats, so \( y=e^{x+y} \) and \( \log y=x+y \).
2. Differentiate: \( \dfrac{y'}y=1+y' \), so \( y'(1-y)=y \) giving \( y'=\dfrac y{1-y} \).

Teacher's Note:
1. Recognise the series or repeated pattern and replace it by y itself.
2. Form an equation in y first and then differentiate that equation implicitly.

Answer: (a) \( \dfrac{y^2\cot x}{1-y\log\sin x} \)

Explanation:
1. The tower repeats, so \( y=(\sin x)^y \) and \( \log y=y\log\sin x \).
2. \( \dfrac{y'}{y}=y'\log\sin x+y\cot x \), hence \( y'(1-y\log\sin x)=y^2\cot x \).
3. \( y'=\dfrac{y^2\cot x}{1-y\log\sin x} \).

Teacher's Note:
1. Form an equation in y first and then differentiate that equation implicitly.
2. Do not differentiate the infinite pattern term by term.

Answer: (d) \( (2y-1)\dfrac{dy}{dx}-\cos x=0 \)

Explanation:
1. The pattern repeats, so \( y=\sqrt{\sin x+y} \) and \( y^2=\sin x+y \).
2. Differentiate: \( 2y\dfrac{dy}{dx}=\cos x+\dfrac{dy}{dx} \), so \( (2y-1)\dfrac{dy}{dx}-\cos x=0 \).

Teacher's Note:
1. Do not differentiate the infinite pattern term by term.
2. Recognise the series or repeated pattern and replace it by y itself.

Answer: (c) \( \frac{1}{x \left(2 y - 1\right)} \)

Explanation:
1. \( y^2=\log x+y \).
2. Differentiate: \( 2yy'=\dfrac1x+y' \), so \( y'=\dfrac{1}{x(2y-1)} \).

Teacher's Note:
1. Recognise the series or repeated pattern and replace it by y itself.
2. Form an equation in y first and then differentiate that equation implicitly.

Answer: (c) \( - \frac{1}{x^{3} y} \)

Explanation:
1. Square the first relation: \( x^4+y^4+2x^2y^2=t^2+2+\dfrac1{t^2} \).
2. Subtract the second relation: \( 2x^2y^2=2 \), so \( x^2y^2=1 \).
3. Differentiate \( x^2y^2=1 \): \( 2xy^2+2x^2y\dfrac{dy}{dx}=0 \), so \( \dfrac{dy}{dx}=-\dfrac yx \).
4. Now \( \dfrac yx=\dfrac{x^2y^2}{x^3y}=\dfrac1{x^3y} \), so \( \dfrac{dy}{dx}=-\dfrac1{x^3y} \).

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) None of these

Explanation:
1. \( \dfrac{dy}{dx}=f'(u)\dfrac{du}{dx} \) with \( u=\dfrac{2x+3}{3-2x} \) and \( \dfrac{du}{dx}=\dfrac{12}{(3-2x)^2} \).
2. So \( \dfrac{dy}{dx}=\dfrac{12}{(3-2x)^2}\sin\left(\log\dfrac{2x+3}{3-2x}\right) \), which matches none of (a), (b), (c).

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) \( y\dfrac{dy}{dx}+x-2y=0 \)

Explanation:
1. Take logs: \( \log(x-y)+\dfrac{x}{x-y}=\log k \).
2. Differentiate: \( \dfrac{1-y'}{x-y}+\dfrac{(x-y)-x(1-y')}{(x-y)^2}=0 \), i.e. \( \dfrac{(1-y')(x-y)+xy'-y}{(x-y)^2}=0 \).
3. Numerator: \( x-y-xy'+yy'+xy'-y=x-2y+yy' \). So \( y\dfrac{dy}{dx}+x-2y=0 \), option (b).

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) \( x(x^x)^x(1+2\log x) \)

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

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) \( \left(1+\dfrac1x\right)^x\left[\log\left(1+\dfrac1x\right)-\dfrac1{1+x}\right] \)

Explanation:
1. \( \log y=x\log\left(1+\dfrac1x\right) \), so \( \dfrac{y'}y=\log\left(1+\dfrac1x\right)+x\cdot\dfrac{-1/x^2}{1+1/x} \).
2. The last term is \( -\dfrac1{x+1} \), so \( y'=\left(1+\dfrac1x\right)^x\left[\log\left(1+\dfrac1x\right)-\dfrac1{1+x}\right] \).

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: (c) \( y[x^x(\log ex)\log x+x^{x-1}] \)

Explanation:
1. \( \log y=x^x\log x \). Since \( \dfrac{d}{dx}x^x=x^x(1+\log x)=x^x\log(ex) \):
2. \( \dfrac{y'}y=x^x\log(ex)\log x+x^x\cdot\dfrac1x=x^x\log(ex)\log x+x^{x-1} \).

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) \( \frac{2}{5} \)

Explanation:
1. \( \dfrac{dx}{dt}=5t^4-15t^2-20 \) and \( \dfrac{dy}{dt}=12t^2-6t-18 \).
2. At \( t=1 \): \( \dfrac{dx}{dt}=5-15-20=-30 \), \( \dfrac{dy}{dt}=12-6-18=-12 \).
3. \( f'(x)=\dfrac{-12}{-30}=\dfrac25 \).

Teacher's Note:
1. Differentiate x and y separately with respect to the parameter and take the ratio.
2. Simplify the ratio with trigonometric identities before substituting the given parameter value.

Answer: (d) None of these

Explanation:
1. The pattern repeats after two steps, so \( y=\dfrac{\sin x}{1+\dfrac{\cos x}{1+y}}=\dfrac{(1+y)\sin x}{1+y+\cos x} \).
2. Hence \( y+y^2+y\cos x-\sin x-y\sin x=0 \).
3. Differentiate: \( y'(1+2y+\cos x-\sin x)=(1+y)\cos x+y\sin x \).
4. So \( y'=\dfrac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x} \), which differs from options (a), (b), (c) (they lack the factor y on \( \sin x \)).

Teacher's Note:
1. Do not differentiate the infinite pattern term by term.
2. Recognise the series or repeated pattern and replace it by y itself.

Answer: (c) \( 1 \)

Explanation:
1. \( f(f(x))=\dfrac{1}{1-\frac1{1-x}}=\dfrac{x-1}{x} \).
2. \( f(f(f(x)))=\dfrac1{1-\frac{x-1}{x}}=x \), so its derivative is 1.

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: (a) \( \dfrac32x\cos x^3\csc x^2 \)

Explanation:
1. \( \dfrac{du}{dx}=f'(x^3)\cdot3x^2=3x^2\cos x^3 \) and \( \dfrac{dv}{dx}=g'(x^2)\cdot2x=2x\sin x^2 \).
2. \( \dfrac{du}{dv}=\dfrac{3x^2\cos x^3}{2x\sin x^2}=\dfrac32x\cos x^3\csc x^2 \).

Teacher's Note:
1. Substitute the point only after you have found the derivative expression.
2. Use the given values of the function and its derivative at the point after applying the rule.

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

Explanation:
1. \( h(x)=e^{\sin^{-1}x} \), so \( \log h=\sin^{-1}x \).
2. \( \dfrac{h'}{h}=\dfrac1{\sqrt{1-x^2}} \).

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: (c) \( 3 x^{3} e^{x^{3}} \)

Explanation:
1. \( \dfrac{d}{dx}e^{x^3}=3x^2e^{x^3} \) and \( \dfrac{d}{dx}\log x=\dfrac1x \).
2. Ratio: \( 3x^3e^{x^3} \).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

Answer: (d) \( - \frac{12}{5} \)

Explanation:
1. \( \dfrac{d}{dx}\sqrt{x^2+16}=\dfrac{x}{\sqrt{x^2+16}}=\dfrac35 \) at \( x=3 \).
2. \( \dfrac d{dx}\dfrac x{x-1}=-\dfrac1{(x-1)^2}=-\dfrac14 \).
3. Ratio \( =\dfrac35\div\left(-\dfrac14\right)=-\dfrac{12}5 \).

Teacher's Note:
1. Simplify both functions with a suitable substitution to get a constant ratio quickly.
2. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.

Answer: (d) None of these

Explanation:
1. Put \( x=\tan^2\theta \): \( \dfrac{1-x}{1+x}=\cos2\theta \), so \( \sin^{-1}\cos2\theta=\dfrac\pi2-2\theta \), and \( \sqrt x=\tan\theta \).
2. \( \dfrac{d(\frac\pi2-2\theta)}{d(\tan\theta)}=\dfrac{-2}{\sec^2\theta}=-\dfrac{2}{1+x} \), which is not among (a), (b), (c).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

Answer: (b) \( 4 \)

Explanation:
1. Put \( x=\cos\theta \): \( \sec^{-1}\dfrac1{\cos2\theta}=2\theta \) and \( \sqrt{1-x^2}=\sin\theta \).
2. \( \dfrac{d(2\theta)}{d(\sin\theta)}=\dfrac{2}{\cos\theta}=\dfrac2x \). At \( x=\dfrac12 \) this is 4.

Teacher's Note:
1. The derivative of u with respect to v is (du/dx) divided by (dv/dx).
2. Simplify both functions with a suitable substitution to get a constant ratio quickly.

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

Explanation:
1. Let \( u=\sqrt x \). Then \( \dfrac{d}{du}\tan^{-1}u=\dfrac1{1+u^2}=\dfrac1{1+x} \).

Teacher's Note:
1. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.
2. The derivative of u with respect to v is (du/dx) divided by (dv/dx).

Answer: (a) \( 0 \)

Explanation:
1. \( \sec^{-1}\dfrac1{2x^2-1}=\cos^{-1}(2x^2-1) \). For \( -1<x<0 \) this equals \( 2\pi-2\cos^{-1}x \), so its derivative is \( \dfrac{2}{\sqrt{1-x^2}} \), which at \( x=-\dfrac13 \) is the finite number \( \dfrac3{\sqrt2} \).
2. \( \dfrac{d}{dx}\sqrt{1+3x}=\dfrac{3}{2\sqrt{1+3x}} \), which becomes infinitely large at \( x=-\dfrac13 \).
3. A finite derivative divided by an infinitely large one gives 0.

Teacher's Note:
1. The derivative of u with respect to v is (du/dx) divided by (dv/dx).
2. Simplify both functions with a suitable substitution to get a constant ratio quickly.

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

Explanation:
1. Let \( \sqrt x=\cos\theta \). Then \( \cos^{-1}\sqrt x=\theta \) and \( \sqrt{1-x}=\sin\theta \).
2. \( \dfrac{d\theta}{d(\sin\theta)}=\dfrac1{\cos\theta}=\dfrac1{\sqrt x} \).

Teacher's Note:
1. Simplify both functions with a suitable substitution to get a constant ratio quickly.
2. If u and v are both multiples of the same angle, the answer is just the ratio of the multiples.