# How do you differentiate  f(x)= (9-x)(5/x^2 -4)  using the product rule?

Dec 24, 2015

The product rule states that for $y = f \left(x\right) g \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

#### Explanation:

Remembering the power rule that states ${a}^{-} n = \frac{1}{{a}^{n}}$, we can rewrite it as follows

$f \left(x\right) = \left(9 - x\right) \left(5 {x}^{-} 2 - 4\right)$

$\left(\mathrm{df} \left(x\right)\right) = \left(- 1\right) \left(5 {x}^{-} 2 - 4\right) + \left(9 - x\right) \left(- 10 {x}^{-} 3\right)$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \left(4 - 5 {x}^{-} 2\right) + \left(- 90 {x}^{-} 3 + 10 {x}^{-} 2\right)$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 4 + 5 {x}^{-} 2 - 90 {x}^{-} 3$

or, if you prefer,

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 4 + \frac{5}{x} ^ 2 - \frac{90}{x} ^ 3$