# Question 0abe4

Jul 30, 2017

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

#### Explanation:

The Quotient Rule States

$\left[\frac{f \left(x\right)}{g \left(x\right)}\right] ' = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g {\left(x\right)}^{2}}$

In this case

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

and

$g \left(x\right) = {x}^{2}$

To find $f ' \left(x\right)$ we will use the Chain Rule, which states

$\left[h \left(k \left(x\right)\right)\right] ' = h ' \left(k \left(x\right)\right) k ' \left(x\right)$

In this case

$h \left(x\right) = \sqrt{x}$

and

$k \left(x\right) = x + 1$

Since the derivative of a constant is $0$ and the derivative of $x$ is $1$

$k ' \left(x\right) = 1$

Also remember that $\sqrt{x} = {x}^{\frac{1}{2}}$

and the Power rule tells us that $\left({x}^{n}\right) ' = n {x}^{n - 1}$

Then

h'(x)=1/2x^(1/2-1)=1/2x^(-1/2)=1/(2sqrt(x)#

Then we plug in and find that

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

and we can use the power rule again to find $g ' \left(x\right)$

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

Then we plug in

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

and we simplify

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

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

$= {x}^{-} \frac{2}{2 \sqrt{x + 1}} - \frac{2 \sqrt{x + 1}}{x} ^ 3$

$= \frac{1}{2 {x}^{2} \sqrt{x + 1}} - \frac{2 \sqrt{x + 1}}{x} ^ 3$