# How do you differentiate f(x)= (x^3+2x+1)(2+x^-2)?

May 19, 2015

This is a simple and delicious case of chain rule, mate!

Let $u = {x}^{3} + 2 x + 1$ and $v = 2 + {x}^{-} 2$ and remember that the chain rule states that

When $y = f \left(x\right) g \left(x\right)$, then $y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$, Let's proceed this way.

$y ' = u ' v + v ' u$

But $u ' = 3 {x}^{2} + 2$ and $v ' = - 2 {x}^{-} 3$

So,

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

Let's just distribute these products in order to try to make it simpler:

$y ' = \left(6 {x}^{2} + 3 + 4 + 2 {x}^{-} 2\right) + \left(- 2 - 4 {x}^{-} 2 - 2 {x}^{-} 3\right)$

Reorganizing,

$y ' = 6 {x}^{2} - 2 {x}^{-} 2 - 2 {x}^{-} 3 + 5$