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

Simplification 

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