# How do you differentiate f(x)=sin(1/(2x)^3) using the chain rule?

May 15, 2017

$f ' \left(x\right) = - \frac{3 \cos \left({\left(2 x\right)}^{- 3}\right)}{8 {x}^{4}}$

#### Explanation:

The derivative of $f , \frac{\mathrm{df}}{\mathrm{dx}} ,$ can be found using the chain rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$

Let $f \left(x\right) = \sin g$ where $g = g \left(x\right) = \frac{1}{2 x} ^ 3 = {\left(2 x\right)}^{-} 3 = \frac{1}{8} {x}^{-} 3$

$\frac{\mathrm{df}}{\mathrm{dg}} = \cos g = \cos \left({\left(2 x\right)}^{-} 3\right)$

$\frac{\mathrm{dg}}{\mathrm{dx}} = - \frac{3}{8} {x}^{-} 4 = - \frac{3}{8 {x}^{4}}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \cos g \times - \frac{3}{8 {x}^{4}}$

$= - \frac{3 \cos \left(\left(2 {x}^{-} 3\right)\right)}{8 {x}^{4}}$