Question

What is the derivative of ${{\tan }^{-1}}\left( 2x \right)$?

Answer 1

Hint: In this problem we need to find the derivative of the given value. For this we are going to use a substitution method and take $u=2x$ as substitution. Now we will consider the substitution and differentiate the equation in order to have the value of $\dfrac{du}{dx}$. After having the value of $\dfrac{du}{dx}$, we will substitute the assumed substitution in the given function and differentiate it with respect to $x$ and use some differentiation formulas like $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$to get the required result.

Complete step by step solution:
Given function is ${{\tan }^{-1}}\left( 2x \right)$.
Considering the substitution $u=2x$ for the above function.
Differentiating the equation $u=2x$ with respect to $x$, then we will have
$\begin{align}
  & \dfrac{du}{dx}=\dfrac{d}{dx}\left( 2x \right) \\
 & \Rightarrow \dfrac{du}{dx}=2\dfrac{dx}{dx} \\
 & \Rightarrow \dfrac{du}{dx}=2 \\
\end{align}$
Substituting the equation $u=2x$ in the given function, then we will get
${{\tan }^{-1}}\left( 2x \right)={{\tan }^{-1}}\left( u \right)$
Differentiating the above equation with respect to $x$, then we will have
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)=\dfrac{d}{dx}\left( {{\tan }^{-1}}u \right)$
Applying the differentiation formula $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ in the above equation, then we will get
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)=\dfrac{1}{1+{{u}^{2}}}\dfrac{du}{dx}$
Substituting the values $u=2x$, $\dfrac{du}{dx}=2$ in the above equation, then we will have
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)=\dfrac{1}{1+{{\left( 2x \right)}^{2}}}\times 2$
Simplifying the above equation by using basic mathematical operations, then we will get
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)=\dfrac{2}{1+4{{x}^{2}}}$

Hence the differentiation value of the given function ${{\tan }^{-1}}\left( 2x \right)$ is $\dfrac{2}{1+4{{x}^{2}}}$.

Note: For this problem we can also use another method which is direct method. In this method we will use the formula $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$. By applying these formulas, we can get the value of $\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)$ as
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)=\dfrac{1}{1+{{\left( 2x \right)}^{2}}}\times \dfrac{d}{dx}\left( 2x \right)$
Simplifying the above equation, then we will get
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( 2x \right) \right)=\dfrac{2}{1+4{{x}^{2}}}$
From both the methods we got the same result.