# How do you differentiate  y =cos(3x+7)  using the chain rule?

Jun 21, 2016

$- 3 \sin \left(3 x + 7\right)$

#### Explanation:

The chain rule: $\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Essentially, you take the derivative of whatever is on the outside like normal then multiply it by whats on the inside of the trig function.

The derivative of $\cos \left(x\right)$ is $- \sin \left(x\right)$
so taking the derivative of the outside gives us: $- \sin \left(3 x + 7\right)$
then we take the derivative of the inside: $\frac{d}{\mathrm{dx}} \left[3 x + 7\right] = 3$

The derivative of the inside is then multiplied by our first derivative:

$- \sin \left(3 x + 7\right) \cdot 3 = - 3 \sin \left(3 x + 7\right)$