# What is the derivative of the following function? Thank you! THANK YOU

Mar 13, 2017

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

#### Explanation:

We can start by simplifying the function.

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

$f \left(x\right) = \frac{x + 1}{x + 2}$

Now, differentiate using the quotient rule.

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

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

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

Hopefully this helps!