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

May 20, 2018

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

#### Explanation:

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$g \left(x\right) = - 4 x \Rightarrow g ' \left(x\right) = - 4$

$h \left(x\right) = {x}^{2} - 1 \Rightarrow h ' \left(x\right) = 2 x$

$\Rightarrow f ' \left(x\right) = \frac{- 4 \left({x}^{2} - 1\right) + 8 {x}^{2}}{{x}^{2} - 1} ^ 2$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{4 {x}^{2} + 4}{{x}^{2} - 1} ^ 2$

May 20, 2018

$y ' = \frac{4 {x}^{2} + 4}{{x}^{2} - 1} ^ 2$

#### Explanation:

show below

$\textcolor{b l u e}{y = \frac{- 4 x}{{x}^{2} - 1}}$

$\textcolor{red}{y = \frac{f \left(x\right)}{g \left(x\right)}}$

The quotient rule $\textcolor{red}{y ' = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)}}$

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

$y ' = \frac{- 4 {x}^{2} + 4 + 8 {x}^{2}}{{x}^{2} - 1} ^ 2 = \frac{4 {x}^{2} + 4}{{x}^{2} - 1} ^ 2$