# How do you differentiate f(x)=tane^(4x) using the chain rule.?

Dec 23, 2015

$f ' \left(x\right) = 4 {e}^{4 x} {\sec}^{2} \left({e}^{4 x}\right)$

#### Explanation:

Treat $f \left(x\right)$ as $\tan \left(u\right)$, where $u = {e}^{4 x}$.

According to the chain rule,

$\frac{d}{\mathrm{dx}} \left[\tan \left(u\right)\right] = {\sec}^{2} \left(u\right) \cdot u '$

So, $f ' \left(x\right) = {\sec}^{2} \left({e}^{4 x}\right) \cdot \frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right]$

To find $\frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right]$, find $\frac{d}{\mathrm{dx}} \left[{e}^{v}\right]$ when $v = 4 x$.

$\frac{d}{\mathrm{dx}} \left[{e}^{v}\right] = {e}^{v} \cdot v '$

$\frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right] = {e}^{4 x} \cdot \frac{d}{\mathrm{dx}} \left[4 x\right]$

$\implies 4 {e}^{4 x}$

Thus,

$f ' \left(x\right) = 4 {e}^{4 x} {\sec}^{2} \left({e}^{4 x}\right)$