# How do you find the derivative for y = ((x-1)/( x+3)) ^ (1/3)?

Jul 30, 2015

You can use the quotient rule and the chain rule.

#### Explanation:

You can differentiate your function by using the chain rule, the quotient rule, and the power rule.

First, you need to recognize that your function can be expressed as

$f \left(u\right) = {u}^{\frac{1}{3}}$, where

$u = \frac{x - 1}{x + 3}$

The chain rule allows you to differentiate a function $y$ that depends on a variable $u$, which in turn depends on a variable $x$ by using the formula

color(blue)(d/dx(y) = d/(du)(y) * d/dx(u)

The derivative of $y$ will thus be equal to

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left({u}^{\frac{1}{3}}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x - 3}\right)$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{1}{3} {u}^{\frac{1}{3} - 1} \cdot \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 3}\right)$

To determine $\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 3}\right)$, you can use the quotient rule, which allows you to differentiate functions that can be expressed as a quotient of two other functions

$a \left(x\right) = \frac{b \left(x\right)}{c \left(x\right)}$

by using the formula

color(blue)(d/dx(a(x)) = (b^'(x) * c(x) - b(x) * c^'(x))/[c(x)]^2, with $c \left(x\right) \ne 0$.

In your case, $\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 3}\right)$ can be differentiated by using the quotient rule to get

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

$\frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 3}\right) = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 3 - \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 1}{x + 3} ^ 2$

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

Plug this into the derivative of $y$ to get

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{1}{3} \cdot {\left(\frac{x - 1}{x + 3}\right)}^{- \frac{2}{3}} \cdot \frac{4}{x + 3} ^ 2$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{4}{3} \cdot \frac{1}{{\left(x - 1\right)}^{\frac{2}{3}} \cdot \frac{1}{x + 3} ^ \left(\frac{2}{3}\right) \cdot {\left(x + 3\right)}^{2}}$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{4}{3} \cdot \frac{1}{{\left(x - 1\right)}^{\frac{2}{3}} \cdot {\left(x + 3\right)}^{\frac{4}{3}}} = \textcolor{g r e e n}{\frac{4}{3} \cdot {\left(x - 1\right)}^{- \frac{2}{3}} \cdot {\left(x + 3\right)}^{- \frac{4}{3}}}$

Alternatively, you can complicate things a bit by writing your original function as

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

and using a combination of the power rule and quotient rule.

This will get a little messy, but you could write

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

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

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

This is equivalent to

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{1}{3} \cdot \left[{\left(x - 1\right)}^{- \frac{2}{3}} \cdot {\left(x + 3\right)}^{- \frac{1}{3}} - {\left(x - 1\right)}^{\frac{1}{3}} \cdot {\left(x + 3\right)}^{- \frac{4}{3}}\right]$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{1}{3} \cdot \left[{\left(x - 1\right)}^{\frac{1}{3}} \cdot {\left(x + 3\right)}^{- \frac{1}{3}} \left[{\left(x - 1\right)}^{- 1} - {\left(x + 3\right)}^{- 1}\right]\right]$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{1}{3} \cdot \left[{\left(x - 1\right)}^{\frac{1}{3}} \cdot {\left(x + 3\right)}^{- \frac{1}{3}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 3 - \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 1}{\left(x - 1\right) \left(x + 3\right)}\right]$

Finally, you have

$\frac{d}{\mathrm{dx}} \left(y\right) = \textcolor{g r e e n}{\frac{4}{3} \cdot {\left(x - 1\right)}^{- \frac{2}{3}} \cdot {\left(x + 3\right)}^{- \frac{4}{3}}}$