# If f(x)= cos(-2 x -1)  and g(x) = 3x^2 -5 , how do you differentiate f(g(x))  using the chain rule?

May 10, 2016

$12 x \sin \left(- 6 {x}^{2} + 9\right)$

#### Explanation:

The chain rule says that the derivative of a compound function like $f \left(g \left(x\right)\right)$ is

$f \left(g \left(x\right)\right) ' = g ' \left(x\right) \cdot f ' \left(g \left(x\right)\right)$,

or, the derivative of the whole thing is the derivative of the inside times by the derivative of the outside (with the inside left the same).

The question gives us the functions, so differentiate them normally, like so

$\frac{d}{\mathrm{dx}} \cos \left(- 2 x - 1\right) = - 2 \cdot - \sin \left(- 2 x - 1\right)$

$= 2 \sin \left(- 2 x - 1\right)$ (this we also get from chain rule, by multiplying the derivative of the inside function by the derivative of the outside, the latter with the inside left alone)

$\frac{d}{\mathrm{dx}} 3 {x}^{2} - 5 = 6 x - 0 = 6 x$

We can put these functions into the chain rule equation,

$f \left(g \left(x\right)\right) ' = g ' \left(x\right) \cdot f ' \left(g \left(x\right)\right)$

$= 6 x \cdot 2 \sin \left(- 2 \cdot g \left(x\right) - 1\right)$

$= 12 x \sin \left(- 2 \left(3 {x}^{2} - 5\right) - 1\right) = 12 x \sin \left(- 6 {x}^{2} + 9\right)$