Chain Rule MCQ Quiz - Objective Question with Answer for Chain Rule - Download Free PDF

Explanation:

Given:

f(0) = -1 and f'(0) = 1 Let h(x) = f(2f(x) +2) and g(x) = (h(x))2 

⇒ g(x) = (h(x))²

⇒ g'(x) = 2(h(x)).h'(x)

⇒ g'(0) = 2(h(0)).h'(0)

= 2f(2f (0) + 2).2

= 2f(0).2 = (–2).2 = –4

∴ Option (a) is correct

Explanation:

Let f: (–1, 1)→ R be a differentiable function with f(0) = –1 and f'(0) = 1

⇒ h(x) = f(2f(x) + 2)

⇒ h'(x) = f'(2f(x) + 2)2f'(x)

⇒ h'(0) = f'(2f(0) + 2).2f'(0)

= f'(–2 + 2).2(1)

= f'(0).2 = (1).2 = 2

∴ Option (d) is correct.

Given:

Differentiate sin(x² + 9) with respect to x.

Formula used:

Chain Rule: If y = sin(u) and u = x² + 9, then dy/dx = cos(u) × du/dx

Calculation:

Let y = sin(x² + 9)

⇒ dy/dx = cos(x² + 9) × (d/dx)(x² + 9)

⇒ dy/dx = cos(x² + 9) × 2x

⇒ dy/dx = 2x cos(x² + 9)

∴ The correct answer is option (4).

Given:

Differentiate sin(x² + 9) with respect to x.

Formula used:

Chain Rule: If y = sin(u) and u = x² + 9, then dy/dx = cos(u) × du/dx

Calculation:

Let y = sin(x² + 9)

⇒ dy/dx = cos(x² + 9) × (d/dx)(x² + 9)

⇒ dy/dx = cos(x² + 9) × 2x

⇒ dy/dx = 2x cos(x² + 9)

∴ The correct answer is option (4).

Concept:

Chain Rule of Derivatives: 

\(\rm \dfrac{d}{dx}f(g(x))=\dfrac{d}{d\ g(x)}f(g(x))\times \dfrac{d}{dx}g(x)\).

\(\rm \dfrac{d}{dx}e^x\) = e^x.

Calculation:

It is given that y = \(\rm e^{x+e^{x+e^{x+^{\ ...\ \infty}}}}\).

∴ y = \(\rm e^{x+(e^{x+e^{x+^{\ ...\ \infty}}})}=e^{x+y}\)

Differentiating both sides with respect to x and using the chain rule, we get:

\(\rm \dfrac{dy}{dx}=\dfrac{d}{dx}e^{x+y}\)

⇒ \(\rm \dfrac{dy}{dx}=e^{x+y}\dfrac{d}{dx}(x+y)\)

⇒ \(\rm \dfrac{dy}{dx}=y\left (1+\dfrac{dy}{dx} \right )\)

⇒ \(\rm \dfrac{dy}{dx}=y+y\dfrac{dy}{dx}\)

⇒ \(\rm (1-y)\dfrac{dy}{dx}=y\)

⇒ \(\rm \dfrac{dy}{dx}=\dfrac{y}{1-y}\).

f'(x) = g(x) and g’(x) = f(x²), then f’’(x²)

given f’(x) = g(x)

Differentiating w.r.t x we get

f’’(x) = g’(x)

f’’(x) = f(x²)

multiply the function with x²

Concept:

Differentiation Formulas

\(\rm \frac{d\: (sin x)}{dx}\) = cos x 

\(\rm \frac{d\: (log \: x)}{dx} = \frac{1}{x}\)

\(\rm \frac{d\: (cos\: x)}{dx} \) = - sin x

Trigonometry Formula

\(\rm tan\: \theta = \frac{sin \: \theta }{cos\: \theta }\)

Calculation:

Given:y = sin (log cos x)

Differentiation with respect to x

\(\rm \frac{dy}{dx}\) = \(\rm \frac{d\;\left \{sin\: (log (cos \:x )) \right \} }{dx}\)

= cos (log (cos x)). \(\rm \frac{d\;\left \{ (log (cos\: x )) \right \} }{dx}\)

= cos (log (cos x)). \(\rm \frac{1}{cos\: x}\) \(\)\(\frac{d\: (cos \:x)}{dx}\)

= cos (log (cos x)). \(\rm \frac{1}{cos\: x}\) .(- sin x)

= - cos (log (cos x)). \(\rm \frac{sin\: x }{cos\: x}\)

= - cos (log (cos x)).tan x​​

\(\rm \frac{dy}{dx}\) = - cos (log (cos x)).tan x

Concept:

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right) = \frac{{{\rm{f'}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{x}} \right){\rm{g'}}\left( {\rm{x}} \right)}}{{{{\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)}^2}}}\)

Calculation:

Given: \({\rm{s}} = \sqrt {{{\rm{t}}^2} + 1} \)

\(\frac{{{\rm{ds}}}}{{{\rm{dt}}}} = \frac{{{\rm{d}}\left( {\sqrt {{{\rm{t}}^2} + 1} } \right)}}{{{\rm{dt}}}}\)

\( = \frac{1}{{2\sqrt {{{\rm{t}}^2} + 1} }} \times \frac{{{\rm{d}}\left( {{{\rm{t}}^2} + 1} \right)}}{{{\rm{dt}}}}\)

\(= \frac{{2{\rm{t}}}}{{2\sqrt {{{\rm{t}}^2} + 1} }} = \frac{{\rm{t}}}{{\sqrt {{{\rm{t}}^2} + 1} }}\)

Now,

\(\frac{{{{\rm{d}}^2}{\rm{s}}}}{{{\rm{d}}{{\rm{t}}^2}}} = \frac{{\rm{d}}}{{{\rm{dt}}}}\left( {\frac{{{\rm{ds}}}}{{{\rm{dt}}}}} \right) = \frac{{\rm{d}}}{{{\rm{dt}}}}\left( {\frac{{\rm{t}}}{{\sqrt {{{\rm{t}}^2} + 1} }}} \right)\)

\(= \frac{{\sqrt {{{\rm{t}}^2} + 1} - \frac{{2{\rm{t}} \times {\rm{t}}}}{{2\sqrt {{{\rm{t}}^2} + 1} }}}}{{{{\left( {\sqrt {{{\rm{t}}^2} + 1} } \right)}^2}}}\)

\( = \frac{{\sqrt {{{\rm{t}}^2} + 1} - \frac{{{{\rm{t}}^2}}}{{\sqrt {{{\rm{t}}^2} + 1} }}}}{{{{\left( {\sqrt {{{\rm{t}}^2} + 1} } \right)}^2}}}\)

\( = \frac{{{{\rm{t}}^2} + 1 - {{\rm{t}}^2}}}{{{{\left( {\sqrt {{{\rm{t}}^2} + 1} } \right)}^3}}}\)

\( = \frac{1}{{{{\left( {\sqrt {{{\rm{t}}^2} + 1} } \right)}^3}}} = \frac{1}{{{{\rm{s}}^3}}}\)

Hence, option (3) is correct.

Concept:​

\(\rm d\over dx\) sin^-1 x = \(\rm 1\over\sqrt{1 - x^2}\)

\(\rm d\over dx\) tan^-1 x = \(\rm 1\over1 + x^2\)

Chain Rule (Differentiation by substitution): If y is a function of u and u is a function of x

\(\rm {dy\over dx} = {dy\over du}× {du\over dx}\)

Calculation:

Let u = 1 - x²

\(\rm du \over dx\) = - 2x

y = sin^-1(1 - x²) = sin^-1 u

\(\rm dy\over dx\) = \(\rm {dy\over du}×{du\over dx}\)            

\(\rm dy\over dx\) = \(\rm d\over du\) sin^-1 u × (-2x)

\(\rm dy\over dx\) = \(\rm 1\over\sqrt{1 - u^2}\) × (-2x)

\(\rm dy\over dx\) = \(\rm -2x\over\sqrt{1 - (1-x^2)^2}\)

\(\rm dy\over dx\) = \(\rm -2x\over\sqrt{2x^2-x^4}\)

\(\rm dy\over dx\) = \(\boldsymbol{\rm -2\over\sqrt{2-x^2}}\)

Concept:

Differentiation Formulas

\(\rm \frac{d\: (sin x)}{dx}\) = cos x 

\(\rm \frac{d}{dx} (e^{x}) = e^{x} \)

\(\rm \frac{d}{dx} (x^{n}) = n x^{n - 1} \)

​

Calculation:

Given:

\(\rm \frac{d}{dx} sin (e^{sin\: 3\sqrt{x}}) \)

= \(\rm cos (e^{sin\: 3\sqrt{x}}) \) \(\rm \frac{d}{dx} (e^{sin\: 3\sqrt{x}}) \)

=  \(\rm cos (e^{sin\: 3\sqrt{x}}) \) \(\rm (e^{sin\: 3\sqrt{x}}) \) \(\rm \frac{d}{dx} ({sin\: 3\sqrt{x}}) \)

=   \(\rm cos (e^{sin\: 3\sqrt{x}}) \) \(\rm (e^{sin\: 3\sqrt{x}}) \) \(\rm \frac{d}{dx} ({sin\: 3\sqrt{x}}) \)

=   \(\rm cos (e^{sin\: 3\sqrt{x}}) \) \(\rm e^{sin\: 3\sqrt{x}}\) \(\rm ({cos\: 3\sqrt{x}}) \) \(\rm \frac{d\: 3\sqrt{x}}{dx} \)

=  ​\(\rm cos (e^{sin\: 3\sqrt{x}}) \)\(\rm e^{sin\: 3\sqrt{x}}\)\(\rm ({cos\: 3\sqrt{x}}) \) . 3 .\(\rm \frac{1}{2\sqrt{x}}\)

= \(\rm \frac{3 \: cos (e^{sin\: 3\sqrt{x}}) e^{sin\: 3\sqrt{x}} ({cos\: 3\sqrt{x}})}{2\sqrt{x}}\)​

Concept:

\(\dfrac{\partial u}{\partial v} = \dfrac{\partial u}{\partial θ} \times \dfrac{\partial θ}{\partial v}\)

Given:

u = \(\rm sin ^{-1}\dfrac{2x}{1+x^2}\)  And v = \(\rm tan ^{-1}\dfrac{2x}{1-x^2}\)

Calculation:

Let,

x = tan θ 

Then, 

u = \(\rm sin ^{-1}\dfrac{2 \tanθ}{1+\tan^2θ}\) = \(\rm \sin ^{-1}{(\sin2θ)}\)

u = 2θ 

\(\dfrac{\partial u}{\partial θ} = \) 2

And,

v = \(\rm tan ^{-1}\dfrac{2x}{1-x^2}\) = \(\rm \tan ^{-1}\dfrac{2\tanθ}{1-\tan^2θ}\) = \(\rm tan ^{-1}{(\tan2θ)}\)

v = 2θ 

\(\dfrac{\partial v}{\partial θ} = \) 2

After that,

\(\dfrac{\partial u}{\partial v} = \dfrac{\partial u}{\partial θ} \times \dfrac{\partial θ}{\partial v}\)

\(\dfrac{\partial u}{\partial v} = 2 \times \dfrac{1}{2}\)

= 1

Concept:

Derivative of sinx with respect to x is cosx

Chain rule:

Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} ⋅ \frac{{dv}}{{dx}}\)

Calculation:

Given function is  y = sin2(2x+5)

We differentiate the function with respect to x

⇒  y' = [sin2(2x+5)]' 

As we know that, \(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} ⋅ \frac{{dv}}{{dx}}\)

⇒ y' = 2 sin(2x+5) ⋅ [sin(2x+5)]'

⇒  y' = 2 sin(2x+5) ⋅ cos(2x+5) ⋅ (2x+5)'

⇒ y' = 2 sin(2x+5).cos(2x+5).(2)

⇒ y' = 2 sin[2(2x+5)]                      (∴ sin2x = 2sinx.cosx)

⇒ y' = 2 sin(4x+10)

Hence, option 4 is correct.

Calculation:

Given:

y = log[log{log(x)}] 

DIfferentiation with respect to x

⇒ \(\rm \frac{{dy}}{{dx}} =\frac {d \;log[log{log(x)}] }{dx}\)     

\(\rm =\frac {d \log[\log{\log(x)}] }{d\log{\log(x)}} \times \frac {d\log{\log(x)}}{dx}\)

\(\rm =\frac {d \log[\log{\log(x)}] }{d\log{\log(x)}} \times \frac {d\log{\log(x)}}{d\log x} \times \frac {d \log x}{dx}\)

\(\rm = \left( {\frac{1}{{\left[ {\log \left\{ {\log \left( x \right)} \right\}} \right]}}} \right) \times \;\left( {\frac{1}{{logx}}} \right) \times \left( {\frac{1}{x}} \right)\) 

\(\rm \frac{{dy}}{{dx}}{\rm{\;}} = \left( {\frac{1}{{\left[ {\left( x \right)\left( {logx} \right)\log \left\{ {\log \left( x \right)} \right\}} \right]}}} \right)\)

Given:

 y = 2x + x log x

Concept:

Use formula

\(\rm \frac{d}{dx}(a^x)=a^x\ loga\)

\(\rm \frac{d}{dx}[f(x)g(x)]=\frac{d}{dx}[f(x)]g(x)+f(x)\frac{d}{dx}[g(x)]\)

Calculation:

 y = 2x + x log x

Differentiate with respect to x

\(\rm \frac{dy}{dx}=2^x\ log 2 + 1\cdot log x + x\cdot\frac{1}{x}\)

\(\rm \frac{dy}{dx}\)= 2x log 2 + log x + 1

Hence the option (4) is correct.