# How do you differentiate  f(x) = (x - 1)^4 /(x^2+2x)^5 using the chain rule?

Nov 16, 2017

${f}^{-} 1 \left(x\right) = \frac{4 {\left(x - 1\right)}^{3} - {\left(x - 1\right)}^{4} \left(10 x + 10\right) {\left({x}^{2} + 2 x\right)}^{3}}{{x}^{2} + 2 x} ^ 6$

#### Explanation:

$\frac{{\left(x - 1\right)}^{4}}{{\left({x}^{2} + 2 x\right)}^{5}}$

First, find which is u and which is v.

$u = {\left(x - 1\right)}^{4}$
$v = {\left({x}^{2} + 2 x\right)}^{5}$

To differentiate both, you have to use the chain rule. Make the 4 the coefficient of the bracket. Then differentiate what is in the bracket, which is 1, and multiply it by 4. Remember to subtract one from the power, becoming 3.

$\frac{\mathrm{du}}{\mathrm{dx}} = 4 \cdot 1 {\left(x - 1\right)}^{3} \to 4 {\left(x - 1\right)}^{3}$

Repeat the same thing for v.

$\frac{\mathrm{dv}}{\mathrm{dx}} = 5 \left(2 x + 2\right) {\left({x}^{2} + 2 x\right)}^{4}$

Now use the quotient rule to find ${f}^{-} 1 \left(x\right)$

quotient rule = $\frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \left({x}^{2} + 2 x\right) {\left(x - 1\right)}^{3} - {\left(x - 1\right)}^{4} \left(10 x + 10\right) {\left({x}^{2} + 2 x\right)}^{4}}{{x}^{2} + 2 x} ^ 7$

${f}^{-} 1 \left(x\right) = \frac{4 {\left(x - 1\right)}^{3} - {\left(x - 1\right)}^{4} \left(10 x + 10\right) {\left({x}^{2} + 2 x\right)}^{3}}{{x}^{2} + 2 x} ^ 6$