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

Jan 17, 2018

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

#### Explanation:

$\text{differentiate using the "color(blue)"chain/product rules}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$\text{differentiate "x/(x-1)^2" using chain/product rules}$

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

$\frac{d}{\mathrm{dx}} \left(x {\left(x - 1\right)}^{2}\right)$

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

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

$\text{differentiate "4/(1-x)" using the chain rule}$

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

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

$\text{returning to the original question}$

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