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

Nov 26, 2015

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

Explanation:

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\left(x - 1\right) {\left(x + 3\right)}^{- 3}\right)$

$= {\left(x + 3\right)}^{- 3} \frac{d}{\mathrm{dx}} \left(x - 1\right) + \left(x - 1\right) \frac{d}{\mathrm{dx}} \left({\left(x + 3\right)}^{- 3}\right)$

$= {\left(x + 3\right)}^{- 3} \left(1\right) + \left(x - 1\right) \left(\left(- 3\right) {\left(x + 3\right)}^{- 4} \left(1\right)\right)$

$= {\left(x + 3\right)}^{- 4} \left(\left(x + 3\right) - 3 \left(x - 1\right)\right)$

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