Question

How do you find the derivative of $ y=\text{arctan}\left( {{x}^{2}} \right) $ ?

Answer 1

Hint: In this problem, we need to calculate the derivative of the given equation. We can observe that the given equation consists of a trigonometric function which is $ \text{arctan} $ which is nothing but an inverse trigonometric function of $ \tan $ . Now we will apply the trigonometric function $ \tan $ on both sides of the given equation and simplify the obtained equation to get the value of $ \tan y $ . After finding the value of $ \tan y $ we will consider it as an equation one and differentiate that equation with respect to $ x $ . Now we will apply all the known formulas of differentiation and calculate the required values to simplify and get the required result.

Complete step by step answer:
Given that, $ y=\text{arctan}\left( {{x}^{2}} \right) $ .
We know that $ \text{arctan} $ is also written as $ {{\tan }^{-1}} $ , then the above equation is modified as
 $ \Rightarrow y={{\tan }^{-1}}\left( {{x}^{2}} \right) $
Applying the trigonometric function $ \tan $ on both sides of the above equation, then we will get
 $ \Rightarrow \tan y=\tan \left( {{\tan }^{-1}}\left( {{x}^{2}} \right) \right) $
We know that when we multiplied a function with its inverse function, then we will get unity as a result. Then we will get
 $ \tan y={{x}^{2}}....\left( \text{i} \right) $
Differentiating the above equation with respect to $ x $ , then we will get
 $ \Rightarrow \dfrac{d}{dx}\left( \tan y \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right) $
We know that $ \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x $ , $ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} $ , then we will get
 $ \begin{align}
  & \Rightarrow {{\sec }^{2}}y\dfrac{dy}{dx}=2x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{{{\sec }^{2}}y} \\
\end{align} $
In the above equation we have the term $ {{\sec }^{2}}y $ . We need to calculate the value of $ {{\sec }^{2}}y $ to complete the differentiation value.
From the equation $ \left( \text{i} \right) $ we have the value of $ \tan y={{x}^{2}} $ . From the trigonometric identity $ {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $ . The value of $ {{\sec }^{2}}y $ is given by
 $ \begin{align}
  & {{\sec }^{2}}y=1+{{\tan }^{2}}y \\
 & \Rightarrow {{\sec }^{2}}y=1+{{\left( {{x}^{2}} \right)}^{2}} \\
 & \Rightarrow {{\sec }^{2}}y=1+{{x}^{4}} \\
\end{align} $
Substituting this value in the differentiation value, then we will get
$ \Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{1+{{x}^{4}}} $.

Note:
We can also directly find the value of $ \dfrac{dy}{dx} $ in another method. We have the equation
 $ y={{\tan }^{-1}}\left( {{x}^{2}} \right) $
Differentiating the above equation with respect to $ x $ , then we will get
 $ \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( {{x}^{2}} \right) \right) $
We have the differentiation formula $ \dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}} $ , then we will have
 $ \dfrac{dy}{dx}=\dfrac{1}{1+{{\left( {{x}^{2}} \right)}^{2}}}\dfrac{d}{dx}\left( {{x}^{2}} \right) $
Applying the formula $ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} $ , then we will get
 $ \dfrac{dy}{dx}=\dfrac{2x}{1+{{x}^{4}}} $
From both the methods we got the same result.