# If f(x)= 1/x  and g(x) = x^3 , how do you differentiate f'(g(x))  using the chain rule?

Nov 1, 2017

$- \frac{3}{x} ^ 4$

#### Explanation:

$f \left(x\right) = \frac{1}{x}$ and $g \left(x\right) = {x}^{3}$

For $f \left(g \left(x\right)\right)$ we use $\frac{1}{g \left(x\right)} = \frac{1}{x} ^ 3$

Derivative of $f \left(g \left(x\right)\right)$ using chain rule:

Let $u = x$

Then.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \left({u}^{-} 3\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} \left(u\right) = - 3 {u}^{-} 4 \cdot 1 = \frac{- 3}{x} ^ 4$