# How do you find the derivative with a square root in the denominator y= 5x/sqrt(x^2+9)?

Aug 4, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10 \left({x}^{2} + 9\right) - 5 x}{\sqrt{{x}^{2} + 9} \left(2 {x}^{2} + 18\right)}$

#### Explanation:

The Quotient Rule says

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

If we let $f \left(x\right) = 5 x$ and $g \left(x\right) = \sqrt{{x}^{2} + 9}$, so that
$f ' \left(x\right) = 5$ and $g ' \left(x\right) = \frac{1}{2} {\left({x}^{2} + 9\right)}^{- \frac{1}{2}}$, and
${\left(g \left(x\right)\right)}^{2} = {x}^{2} + 9$. Plugging these results in the Quotient Rule, we then have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{5 \sqrt{{x}^{2} + 9} - 5 x \frac{1}{2} {\left({x}^{2} + 9\right)}^{- \frac{1}{2}}}{{x}^{2} + 9}$
$= \left(\frac{2 \times}{2 \times}\right) \frac{5 \sqrt{{x}^{2} + 9} - 5 x \frac{1}{2} {\left({x}^{2} + 9\right)}^{- \frac{1}{2}}}{{x}^{2} + 9}$
$= \left(\frac{\sqrt{{x}^{2} + 9} \times}{\sqrt{{x}^{2} + 9} \times}\right) \frac{10 \sqrt{{x}^{2} + 9} - 5 x {\left({x}^{2} + 9\right)}^{- \frac{1}{2}}}{2 {x}^{2} + 18}$
$= \frac{10 \left({x}^{2} + 9\right) - 5 x}{\sqrt{{x}^{2} + 9} \left(2 {x}^{2} + 18\right)}$