# How do you differentiate f(x)=sqrt((x^3+x^2)^-1+x^3)  using the chain rule?

Jul 13, 2018

$f ' \left(x\right) = \frac{- {\left({x}^{3} + {x}^{2}\right)}^{- 2} \left(3 {x}^{2} + 2 x\right) + 3 {x}^{2}}{\sqrt{{\left({x}^{3} + {x}^{2}\right)}^{- 1} + {x}^{3}}}$

#### Explanation:

We Need the chain rule

(f(g(x))'=f'(g(x))*g'(x)

so we get

$f ' \left(x\right) = \frac{1}{2} \cdot {\left({\left({x}^{3} + {x}^{2}\right)}^{- 1} + {x}^{3}\right)}^{- \frac{1}{2}} \cdot \left(- {\left({x}^{3} + {x}^{2}\right)}^{- 2} \left(3 {x}^{2} + 2 x\right) + 3 {x}^{2}\right)$

this is
$\frac{\left(- {\left({x}^{3} + {x}^{2}\right)}^{- 2} \left(3 {x}^{2} + 2 x\right) + 3 {x}^{2}\right)}{\sqrt{{\left({x}^{3} + {x}^{2}\right)}^{- 1} + {x}^{3}}}$