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

Oct 7, 2016

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

Explanation:

The product rule is $\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$; so
$f ' \left(x\right) = \left({x}^{3} + 2 x + 1\right) \frac{d}{\mathrm{dx}} \left(2 + {x}^{-} 2\right) + \frac{d}{\mathrm{dx}} \left({x}^{3} + 2 x + 1\right) \left(2 + {x}^{-} 2\right)$

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

$f ' \left(x\right) = \left(- 2\right) \frac{{x}^{3} + 2 x + 1}{{x}^{3}} + \left(6 {x}^{2} + 3 {x}^{-} 2 + 4 + 2 {x}^{-} 2\right)$

$f ' \left(x\right) = \left(- 2\right) \left(1 + 2 {x}^{-} 2 + {x}^{-} 3\right) + \left(6 {x}^{2} + 4 + 5 {x}^{-} 2\right)$

$f ' \left(x\right) = - 2 - 4 {x}^{-} 2 - 2 {x}^{-} 3 + 6 {x}^{2} + 4 + 5 {x}^{-} 2$

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