# How do you differentiate f(x)= (x-1)/( x+3) ^ (1/3  using the quotient rule?

May 1, 2017

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

#### Explanation:

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

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

$\text{also require to differentiate using the "color(blue)"chain rule}$

d/dx(f(g(x)))=f'(g(x)xxg'(x)

$\text{here } g \left(x\right) = x - 1 \Rightarrow g ' \left(x\right) = 1$

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

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

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

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