How do you differentiate f(x) = 5/(x^3+4) using the quotient rule?

Jan 23, 2016

$f ' \left(x\right) = \frac{- 15 {x}^{2}}{{x}^{3} + 4} ^ 2$

Explanation:

Application of the quotient rule shows that

$f ' \left(x\right) = \frac{\left({x}^{3} + 4\right) \frac{d}{\mathrm{dx}} \left[5\right] - 5 \frac{d}{\mathrm{dx}} \left[{x}^{3} + 4\right]}{{x}^{3} + 4} ^ 2$

Find each derivative.

$f ' \left(x\right) = \frac{\left({x}^{3} + 4\right) \left(0\right) - 5 \left(3 {x}^{2}\right)}{{x}^{3} + 4} ^ 2$

$f ' \left(x\right) = \frac{- 15 {x}^{2}}{{x}^{3} + 4} ^ 2$