# What is the derivative of y= (5x)/sqrt (x^2+9)?

Apr 13, 2017

#### Answer:

$y ' = \frac{45}{{x}^{2} + 9} ^ \left(\frac{3}{2}\right)$

#### Explanation:

By rewriting a bit,

$y = 5 \frac{x}{{x}^{2} + 9} ^ \left(\frac{1}{2}\right)$

$y ' = 5 \frac{1 \cdot {\left({x}^{2} + 9\right)}^{\frac{1}{2}} - x \cdot \frac{1}{\cancel{2}} {\left({x}^{2} + 9\right)}^{- \frac{1}{2}} \left(\cancel{2} x\right)}{{\left({x}^{2} + 9\right)}^{\frac{1}{2}}} ^ 2$

By cleaning up a bit,

$= 5 \frac{{\left({x}^{2} + 9\right)}^{\frac{1}{2}} - {x}^{2} / {\left({x}^{2} + 9\right)}^{\frac{1}{2}}}{{x}^{2} + 9}$

By multiply the numerator and the denominator by ${\left({x}^{2} + 9\right)}^{\frac{1}{2}}$,

$= 5 \frac{\cancel{{x}^{2}} + 9 - \cancel{{x}^{2}}}{{x}^{2} + 9} ^ \left(\frac{3}{2}\right) = \frac{45}{{x}^{2} + 9} ^ \left(\frac{3}{2}\right)$

I hope that this was clear.