# How do you differentiate f(x)=sin(4-x^2)  using the chain rule?

$f ' \left(x\right) = - 2 x \cdot \cos \left(4 - {x}^{2}\right)$

#### Explanation:

The solution :

The formula for differentiating $\sin u$ for any differentiable function $u$ is

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

The given: $f \left(x\right) = \sin \left(4 - {x}^{2}\right)$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(4 - {x}^{2}\right)\right) = \cos \left(4 - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right)$

$f ' \left(x\right) = \cos \left(4 - {x}^{2}\right) \cdot \left(- 2 x\right)$

$f ' \left(x\right) = - 2 x \cdot \cos \left(4 - {x}^{2}\right)$

God bless....I hope the explanation is useful.