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This is one of our many quizzes on Quotient rule. Click on this to access more such competitive quizzes on differentiation.

Quotient rule Quiz 2

mathematics

 Value of derivative of $\dfrac{{{x}^{2}}+1}{{{x}^{2}}-1}$ is

 Use quotient rule to solve this question.

$\begin{align} &nbsp;&nbsp;& f(x)=\dfrac{{{x}^{2}}+1}{{{x}^{2}}-1} \\ &nbsp;& \text{Apply quotient rule }\dfrac{u}{v}=\dfrac{v.{{u}^{'}}-u.{{v}^{'}}}{{{v}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{\left( {{x}^{2}}-1 \right)\left( 2x \right)-\left( {{x}^{2}}+1 \right)\left( 2x \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{2{{x}^{3}}-2x-2{{x}^{3}}-2x}{{{\left( {{x}^{2}}-1 \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{-4x}{{{\left( {{x}^{2}}-1 \right)}^{2}}} \\ &nbsp;& \text{Hence }f'\left( x \right)=\dfrac{-4x}{{{\left( {{x}^{2}}-1 \right)}^{2}}} \\ \end{align}$

 Value of derivative of $\dfrac{\tan x}{\cos ecx}$ is

 Apply quotient rule in this question as terms are in division and later use chain rule to simplify.

$\begin{align} &nbsp;&nbsp;& f\left( x \right)=\dfrac{\tan x}{\cos ecx}=\dfrac{\dfrac{\sin x}{\operatorname{cosx}}}{\dfrac{1}{\sin x}} \\ &nbsp;& f\left( x \right)=\dfrac{{{\sin }^{2}}x}{\cos x} \\ &nbsp;& \text{Apply quotient rule }\dfrac{u}{v}=\dfrac{v.{{u}^{'}}-u.{{v}^{'}}}{{{v}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{\cos x\left( 2\sin x\cos x \right)-\left( {{\sin }^{2}}x \right){{\left( \cos x \right)}^{'}}}{{{\cos }^{2}}x} \\ &nbsp;& \Rightarrow \dfrac{\cos x\left( 2\sin x\cos x \right)-\left( {{\sin }^{2}}x \right)\left( -\sin x \right)}{{{\cos }^{2}}x} \\ &nbsp;& \Rightarrow \dfrac{\cos x.\sin 2x+\left( {{\sin }^{3}}x \right)}{{{\cos }^{2}}x}\left[ 2\sin x\cos x=\sin 2x \right] \\ &nbsp;& f'\left( x \right) = \dfrac{\cos x.\sin 2x+\left( {{\sin }^{3}}x \right)}{{{\cos }^{2}}x} \\ \end{align}$

 Value of derivative of $\dfrac{{{e}^{x}}}{1+\cos x}$ is

 We need to apply quotient rule in this question as terms are given in both numerator and denominator.

$\begin{align} &nbsp;&nbsp;& f\left( x \right)=\dfrac{{{e}^{x}}}{1+\cos x} \\ &nbsp;& \text{Quotient rule }\dfrac{u}{v}=\dfrac{v.{{u}^{'}}-u.{{v}^{'}}}{{{v}^{2}}} \\ &nbsp;& f'\left( x \right)=\dfrac{(1+\cos x){{\left( {{e}^{x}} \right)}^{'}}-\left( {{e}^{x}} \right){{\left( 1+\cos x \right)}^{'}}}{{{\left( 1+\cos x \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{\left( {{e}^{x}} \right)(1+\cos x)-\left( {{e}^{x}} \right)\left( -\sin x \right)}{{{\left( 1+\cos x \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{{{e}^{x}}+{{e}^{x}}\cos x+{{e}^{x}}\sin x}{{{\left( 1+\cos x \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{{{e}^{x}}\left( \cos x+\sin x \right)+{{e}^{x}}}{{{\left( 1+\cos x \right)}^{2}}} \\ &nbsp;& \Rightarrow f'\left( x \right) = \dfrac{{{e}^{x}}\left( \cos x+\sin x \right)+{{e}^{x}}}{{{\left( 1+\cos x \right)}^{2}}} \\ \end{align}$

 Value of derivative of $\dfrac{{{e}^{x}}}{\tan x}$ is

$\begin{align} &nbsp;&nbsp;& f\left( x \right)=\dfrac{{{e}^{x}}}{\tan x} \\ &nbsp;& \text{Quotient rule }\dfrac{u}{v}=\dfrac{v.{{u}^{'}}-u.{{v}^{'}}}{{{v}^{2}}} \\ &nbsp;& f'(x)\Rightarrow \dfrac{\tan x{{\left( {{e}^{x}} \right)}^{'}}-\left( {{e}^{x}} \right){{\left( \tan x \right)}^{'}}}{{{\tan }^{2}}x} \\ &nbsp;& \Rightarrow \dfrac{\tan x{{e}^{x}}-{{e}^{x}}{{\sec }^{2}}x}{{{\tan }^{2}}x} \\ &nbsp;& \Rightarrow \dfrac{{{e}^{x}}(\tan x-{{\sec }^{2}}x)}{{{\tan }^{2}}x} \\ &nbsp;& \text{Hence, }f'\left( x \right) = \dfrac{{{e}^{x}}(\tan x-{{\sec }^{2}}x)}{{{\tan }^{2}}x} \\ \end{align}$

 Value of $F'\left( x \right)$, if $f\left( x \right)=\dfrac{2{{x}^{2}}}{2x-3}$ is

 We need to apply quotient rule here as the terms are given in both numerator and denominator. 

$\begin{align} &nbsp;&nbsp;& f\left( x \right)=\dfrac{2{{x}^{2}}}{2x-3} \\ &nbsp;& \text{Quotient rule }\dfrac{u}{v}=\dfrac{v.{{u}^{'}}-u.{{v}^{'}}}{{{v}^{2}}} \\ &nbsp;& f'\left( x \right)=\dfrac{\left( 2x-3 \right){{\left( 2{{x}^{2}} \right)}^{'}}-\left( 2{{x}^{2}} \right){{\left( 2x-3 \right)}^{'}}}{{{\left( 2x-3 \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{\left( 2x-3 \right)\left( 4x \right)-\left( 2{{x}^{2}} \right)\left( 2 \right)}{{{\left( 2x-3 \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{8{{x}^{2}}-12x-4{{x}^{2}}}{{{\left( 2x-3 \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{4{{x}^{2}}-12x}{{{\left( 2x-3 \right)}^{2}}} \\ &nbsp;& \Rightarrow \dfrac{4x\left( x-3 \right)}{{{\left( 2x-3 \right)}^{2}}} \\ &nbsp;& \Rightarrow f'\left( x \right) = \dfrac{4x\left( x-3 \right)}{{{\left( 2x-3 \right)}^{2}}} \\ \end{align}$