# How do you find the derivative for y= (x^2 + 2x + 5) / (x + 1)?

Jul 31, 2015

${y}^{'} = 1 - \frac{4}{x + 1} ^ 2$

#### Explanation:

You could differentiate this function by using the quotient rule, or you could simplify the function first, then use the chain rule.

Here's how you could simplify this function. Notice that the numerator can be written as

$y = \frac{{x}^{2} + 2 x + 1 + 4}{x + 1}$

This is an important thing to notice because you can use the formula for the square of a binomial

$\textcolor{b l u e}{{\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}}$

to write

$y = \frac{{\left(x + 1\right)}^{2} + 4}{x + 1}$

You can then go ahead and simplify this by

$y = {\left(x + 1\right)}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} / \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 1\right)}}}\right) + \frac{4}{x + 1}$

$y = x + 1 + \frac{4}{x + 1}$This can be differentiated by using the chain rule for

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

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

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

Your target derivative will thus be equal to

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

${y}^{'} = 1 + 0 - 4 \cdot \frac{1}{x + 1} ^ 2 = \textcolor{g r e e n}{1 - \frac{4}{x + 1} ^ 2}$