# How do you differentiate f(x)=(x^2+4x)*(2+x^-2) using the product rule?

Feb 18, 2016

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

#### Explanation:

$f ' \left(x\right) = \frac{\text{d"}{"d} x}{\left({x}^{2} + 4 x\right) \cdot \left(2 + {x}^{- 2}\right)}$

$= \left({x}^{2} + 4 x\right) \cdot \frac{\text{d"}{"d} x}{2 + {x}^{- 2}}$

$+ \left(2 + {x}^{- 2}\right) \cdot \frac{\text{d"}{"d} x}{{x}^{2} + 4 x}$

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

$+ \left(2 + {x}^{- 2}\right) \cdot \left(2 x + 4\right)$

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

$= 4 x + 8 - \frac{4}{x} ^ 2$

Feb 18, 2016

$- 2 {x}^{-} 2 \left(x + 4\right) + 2 \left(2 + {x}^{-} 2\right) \left(x + 2\right)$

#### Explanation:

using the $\textcolor{b l u e}{\text{ product rule }}$

If f(x) = g(x).h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)

hence $f ' \left(x\right) = \left({x}^{2} + 4 x\right) \frac{d}{\mathrm{dx}} \left(2 + {x}^{-} 2\right) + \left(2 + {x}^{-} 2\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 4 x\right)$

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

$= x \left(x + 4\right) \left(- 2 {x}^{-} 3\right) + \left(2 + {x}^{-} 2\right) 2 \left(x + 2\right)$
(removed 'common factors' from the 2 brackets of x and 2)

$\Rightarrow = - 2 {x}^{-} 2 \left(x + 4\right) + 2 \left(2 + {x}^{-} 2\right) \left(x + 2\right)$