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

Aug 29, 2015

${y}^{'} = \frac{6 \left(x - 1\right) \left(x + 1\right)}{{x}^{2} + 3} ^ 3$

#### Explanation:

One way in which you can differentiate this function is by using the product rule and the chain rule.

You know that for a function $f \left(x\right)$ that can be written as the product of two other functions, $g \left(x\right)$ and $h \left(x\right)$, its derivative can be calculated using

color(blue)(d/dx(f(x)) = [d/dx(g(x))] * h(x) + g(x) * d/dx(h(x))

$g \left(x\right) = - 2 x \text{ }$ and $\text{ } h \left(x\right) = {\left({x}^{2} + 3\right)}^{- 2}$

This means that you have

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(- 2 x\right)\right] \cdot {\left({x}^{2} + 3\right)}^{- 2} + \left(- 2 x\right) \cdot \frac{d}{\mathrm{dx}} {\left({x}^{2} + 3\right)}^{- 2}$

To differentiate ${\left({x}^{2} + 3\right)}^{- 2}$, use the chain rule for ${u}^{- 2}$, with $u = {x}^{2} + 3$. This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{- 2}\right) = \frac{d}{\mathrm{du}} \left({u}^{- 2}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({u}^{- 2}\right) = - 2 {u}^{- 3} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 3\right)$

d/dx*u^(-2)) = -2(x^2 + 3)^(-3) * 2x

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

Take this back to your target derivative

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

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

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

${y}^{'} = \frac{6 \left({x}^{2} - 1\right)}{{x}^{2} + 3} ^ 3 = \textcolor{g r e e n}{\frac{6 \left(x - 1\right) \left(x + 1\right)}{{x}^{2} + 3} ^ 3}$