# How do you differentiate f(x)=cos(7-4x)  using the chain rule?

##### 1 Answer
Mar 24, 2016

$f ' \left(x\right) = 4 \sin \left(7 - 4 x\right)$

#### Explanation:

The chain rule, when applied to $\cos \left(x\right)$, states that

$\frac{d}{\mathrm{dx}} \cos \left(u\right) = - \sin \left(u\right) \cdot u '$

This is very similar to the typical differentiation for $\cos \left(x\right)$ without the chain rule:

$\frac{d}{\mathrm{dx}} \cos \left(x\right) = - \sin \left(x\right)$

except for that when the chain rule is applied the derivative of the function $u$ is also multiplied to the $- \sin \left(u\right)$ expression.

Applying this to $f \left(x\right) = \cos \left(7 - 4 x\right)$, where $u = 7 - 4 x$, we see that

$f ' \left(x\right) = - \sin \left(7 - 4 x\right) \cdot \frac{d}{\mathrm{dx}} \left(7 - 4 x\right)$

Note that the derivative of $7 - 4 x$ is $- 4$. This simplifies our overall derivative expression:

$f ' \left(x\right) = - \sin \left(7 - 4 x\right) \cdot \left(- 4\right)$

$f ' \left(x\right) = 4 \sin \left(7 - 4 x\right)$