# How do you use the quotient rule to differentiate f(x)= (-4x)/ (x^2 - 1) ^2?

Mar 8, 2016

$\frac{12 {x}^{2} + 4}{{x}^{2} - 1} ^ 3$

#### Explanation:

Using the $\textcolor{b l u e}{\text{ Quotient rule }}$

If $f \left(x\right) = g \frac{x}{h \left(x\right)} \text{ then } f ' \left(x\right) = \frac{h \left(x\right) . g ' \left(x\right) - g \left(x\right) . h ' \left(x\right)}{h {\left(x\right)}^{2}}$

and$\textcolor{red}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$
$\text{-------------------------------------------------------------------------}$

here g(x) = - 4x → g'(x) = - 4

and h(x) $= {\left({x}^{2} - 1\right)}^{2} \Rightarrow h ' \left(x\right) = 2 \left({x}^{2} - 1\right) \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right) = 4 x \left({x}^{2} - 1\right)$

substitute these results into f'(x) above:

$f ' \left(x\right) = \frac{{\left({x}^{2} - 1\right)}^{2.} \left(- 4\right) - \left(- 4 x\right) .4 x \left({x}^{2} - 1\right)}{{x}^{2} - 1} ^ 4$

$= \frac{- 4 {\left({x}^{2} - 1\right)}^{2} + 16 {x}^{2} \left({x}^{2} - 1\right)}{{x}^{2} - 1} ^ 4$

$= \frac{\left({x}^{2} - 1\right) \left[- 4 \left({x}^{2} - 1\right) + 16 {x}^{2}\right)}{{x}^{2} - 1} ^ 4$

$= \frac{\cancel{{x}^{2} - 1} \left(12 {x}^{2} + 4\right)}{\cancel{{x}^{2} - 1} {\left({x}^{2} - 1\right)}^{3}} = \frac{12 {x}^{2} + 4}{{x}^{2} - 1} ^ 3$