Question

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

Answer 1

Hint : In order to derivative the value given use chain rule as let $\tan \left( {{x^2}} \right)$ as $u$ which becomes $u = \tan \left( {{x^2}} \right)$ and where ${x^2}$ is considered as $v$ which gives $u = \tan v$ . Now, using Chain rule derivative $u$ with respect to $v$ and $v$ with respect to $x$ .

Complete step by step solution:
We are given the equation $\tan \left( {{x^2}} \right)$ .
We will use the chain rule to solve the above equation. For that write $\tan \left( {{x^2}} \right)$ as $u$ which gives $u = \tan \left( {{x^2}} \right)$ and where ${x^2}$ is considered as $v$ which gives $u = \tan v$ .
Now, form the chain rule, derivative $u$ with respect to $v$ and $v$ with respect to $x$ .
First, derivate both sides of $u$ with respect to $v$ as $u$ has a variable $v$ and we get:
$u = \tan v \\\ \dfrac{{du}}{{dv}} = \dfrac{{d(\tan v)}}{{dv}} = {\sec ^2}v \;$
Then derivative both sides of $v$ with respect to ${x^2}$ as $v$ has a variable $x$ and we get:
$v = {x^2} \\\ \dfrac{{dv}}{{dx}} = \dfrac{{d{x^2}}}{{dx}} = 2x \;$
Combine the values as obtained and we get:
$\dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}} = {\sec ^2}v \times 2x \\\ \dfrac{{du}}{{dx}} = 2x.{\sec ^2}v \;$
where, ${\sec ^2}v$ can be written as ${\sec ^2}\left( {{x^2}} \right)$ .
That implies $\dfrac{{du}}{{dx}} = 2x{\sec ^2}\left( {{x^2}} \right)$ .
Hence, the derivative of $\tan \left( {{x^2}} \right)$ is $2x.{\sec ^2}\left( {{x^2}} \right)$ .
So, the correct answer is “ $2x.{\sec ^2}\left( {{x^2}} \right)$ ”.

Note : Chain rule is a method to split the equation in different parts with different variables which help us to solve the equation easily by derivating each part at once, then collectively combining them to get the required results.
We can solve the above equation directly if we are confident of chain rule without doing step by step but it’s always preferred to do step by step because when there are more than two terms or variables then it becomes difficult to solve directly.
It's very important to remember small formula’s, otherwise there can be a chance of error with wrong formula insertion.