# What is the derivative of sqrt(x)/(x^3+1) using the quotient rule?

Apr 11, 2015

We'll use the quotient rule, and, because the square root is used so often (for example: diagonal distance = hypotenuse of a right triangle), it's nice to memorize:

$\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2 \sqrt{x}}$ (Go through it witl exponents to see that it's true.)

So:

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

Mulhtiply by $\frac{2 \sqrt{x}}{2 \sqrt{x}}$ to get:

$\frac{d}{\mathrm{dx}} \left(\frac{\sqrt{x}}{{x}^{3} + 1}\right) = \frac{\left({x}^{3} + 1\right) - 2 x \left(3 {x}^{2}\right)}{\left(2 \sqrt{x}\right) {\left({x}^{3} + 1\right)}^{2}} = \frac{- 5 {x}^{2} + 1}{\left(2 \sqrt{x}\right) {\left({x}^{3} + 1\right)}^{2}}$