Question

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

Answer 1

Solve for differentiation:

Given: $\frac{dy}{dx}=\frac{d}{dx}\left({\tan }^2\left(x\right)\right)$

$y=f\left(g\right(x\left)\right)$
applying chain rule $\frac{df\left(g\right(x\left)\right)}{dx}=\frac{d}{d\left(g\right(x\left)\right)}f\left(g\right(x\left)\right)\times \frac{d}{dx}g\left(x\right)$

Here $f\left(g\right(x\left)\right)={\left[\tan \left(x\right)\right]}^2$ and $g\left(x\right)=\tan \left(x\right)$

$\frac{d}{dx}(\tan {\left(x\right))}^2=\frac{d}{d\left(\tan \right(x\left)\right)}(\tan {\left(x\right))}^2\times \frac{d}{dx}\left(\tan \right(x\left)\right)$

$$\begin{gathered} \frac{d}{dx}\left({\tan }^2\right(x\left)\right)=2\tan \left(x\right)\times \frac{d}{dx}\left(\tan x\right) \\ \frac{dy}{dx}=2\tan \left(x\right)\times {\sec }^2\left(x\right)\left[\frac{d}{dx}\tan \left(x\right)={\sec }^2\left(x\right)\right] \\ \frac{dy}{dx}=2{\sec }^2\left(x\right)\tan \left(x\right) \end{gathered}$$

Hence, the derivative of $y={\tan }^2\left(x\right)$ is $\frac{dy}{dx}=2{\sec }^2\left(x\right)\tan \left(x\right)$.