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

Mar 23, 2016

$f ' \left(x\right) = - 4 {e}^{4 x} \csc \left({e}^{4 x}\right) \cot \left({e}^{4 x}\right)$

#### Explanation:

The chain rule states that when differentiating a function inside of a function, (1) differentiate the outside function and leave the inside function as is, and (2) multiply this by the derivative of the inside version.

In $f \left(x\right) = \csc \left({e}^{4 x}\right)$, we have the outside function $\csc \left(u\right)$ and the inside function ${e}^{4 x}$.

Thus, since the derivative of $\csc \left(x\right)$ is $- \csc \left(x\right) \cot \left(x\right)$, we will have

$f ' \left(x\right) = - \csc \left({e}^{4 x}\right) \cot \left({e}^{4 x}\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{4 x}\right)$

Don't forget to multiply by the derivative of the inside function, which is $\frac{d}{\mathrm{dx}} \left({e}^{4 x}\right)$.

To differentiate ${e}^{4 x}$, we will have to use the chain rule again.

The outside function is ${e}^{u}$ and the inside function is $u = 4 x$. Since the derivative of ${e}^{x}$ is still ${e}^{x}$, we have

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

Note that $\frac{d}{\mathrm{dx}} \left(4 x\right) = 4$, so we know that

$\frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) = 4 {e}^{4 x}$

Plug this back in to the derivative of the whole function:

$f ' \left(x\right) = - 4 {e}^{4 x} \csc \left({e}^{4 x}\right) \cot \left({e}^{4 x}\right)$