Question

Fill in the blank:

$\frac{d}{dx}\left(\tan hx\right)=\_\_\_\_\_\_\_\_\_.$

Short answer

$\frac{d}{dx}\left(\tan hx\right)=sec{h}^{2}x.$

Step-by-step solution

Given Information 

The given expression$\frac{d}{dx}\left(\tan hx\right).$

Simplification 

$$\begin{gathered} \frac{d}{dx}\left(\tan hx\right) \\ =\frac{d}{dx}(\frac{\sin hx)}{\cos hx)} \\ =\frac{{\displaystyle \frac{d}{dx}}\left(\sin hx\right).\cos hx-\sin hx{\displaystyle \frac{d\left(\cos hx\right)}{dx}}}{\cos h^{2}x} \\ =\frac{\cos hx.\cos hx-\sin hx.\sin hx}{\cos h^{2}x} \\ =\frac{cos{h}^{2}x-\sin h^{2}x}{\cos h^{2}x} \\ =\frac{1}{\cos h^{2}x} \\ =sec{h}^{2}x \\ Hence,\frac{d}{dx}(\tan hx)=sec{h}^{2}x. \end{gathered}$$