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

Jul 22, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt[3]{x + 3}} - \frac{x - 1}{3 {\left(x + 3\right)}^{\frac{4}{3}}}$

#### Explanation:

I'm assuming you mean differentiate? I don't think "derive" is the correct word in this context. The quotient rule is given by:

$\frac{d}{\mathrm{dx}} \left(\frac{f \left(x\right)}{g \left(x\right)}\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

We have $f \left(x\right) = x - 1 \implies f ' \left(x\right) = 1$

and

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

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

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

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

Tidying up and removing negative exponents:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt[3]{x + 3}} - \frac{x - 1}{3 {\left(x + 3\right)}^{\frac{4}{3}}}$