# How do you find the derivative of -2x(x^2+3)^-2?

Jun 28, 2016

A combination of the chain rule and product rule will give you the derivative of this function.

Let your function be $f \left(x\right) = g \left(x\right) \times h \left(x\right)$

The derivative is given by $f ' \left(x\right) = g ' \left(x\right) \times h \left(x\right) + h ' \left(x\right) \times g \left(x\right)$

The derivative of $g \left(x\right)$ is simple enough: $g ' \left(x\right) = - 2$

However, we will need the chain rule to differentiate $h \left(x\right)$ .

$y = {u}^{- 2}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{{u}^{3}} \times 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 x}{{x}^{2} + 3} ^ 3$

Now, let's use the product rule, as mentioned above, to determine the derivative of $f \left(x\right)$.

$f ' \left(x\right) = g ' \left(x\right) \times h \left(x\right) + h ' \left(x\right) \times g \left(x\right)$

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

$f ' \left(x\right) = - \frac{2}{{x}^{2} + 3} ^ 2 + \frac{8 {x}^{2}}{{x}^{2} + 3} ^ 3$

$f ' \left(x\right) = \frac{8 {x}^{2}}{{x}^{2} + 3} ^ 3 - \frac{2}{{x}^{2} + 3} ^ 2$

Hence, the derivative of $y = - 2 x {\left({x}^{2} + 3\right)}^{- 2}$ is $y ' = \frac{8 {x}^{2}}{{x}^{2} + 3} ^ 3 - \frac{2}{{x}^{2} + 3} ^ 2$

Practice exercises:

1. Differentiate the following functions.

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

b) $g \left(x\right) = 2 {x}^{2} \sqrt{{x}^{5} - 9}$

c) $h \left(x\right) = 4 {x}^{3} \sqrt[3]{5 {x}^{2} - 7}$

d) $i \left(x\right) = 2 {x}^{4} {\left(4 {x}^{3} - 11 x\right)}^{-} 14$

Hopefully this helps, and good luck!