# How do you find the derivative for sqrt(x)/(x^3+1)?

Aug 19, 2015

${y}^{'} = \frac{1 - 5 {x}^{3}}{2 \sqrt{x} \cdot {\left({x}^{3} + 1\right)}^{2}}$

#### Explanation:

You can differentiate this function by using the quotient rule, which allows you to differentiate a function that can be written as

$\textcolor{b l u e}{y = f \frac{x}{g} \left(x\right)}$, with $g \left(x\right) \ne 0$

by using the formula

color(blue)(d/dx(y) = ([d/dx(f(x))] * g(x) - f(x) * d/dx(g(x)))/((g(x))^2)

In your case, you can say that

$f \left(x\right) = \sqrt{x} \text{ }$ and $\text{ } g \left(x\right) = {x}^{3} + 1$

This means that you can write

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

${y}^{'} = \frac{\frac{1}{2} \cdot {x}^{- \frac{1}{2}} \cdot \left({x}^{3} + 1\right) - {x}^{\frac{1}{2}} \cdot 3 {x}^{2}}{{x}^{3} + 1} ^ 2$

${y}^{'} = \frac{{x}^{- \frac{1}{2}} \cdot \left[\frac{1}{2} \cdot \left({x}^{3} + 1\right) - x \cdot 3 {x}^{2}\right]}{{x}^{3} + 1} ^ 2$

${y}^{'} = \frac{{x}^{- \frac{1}{2}} \cdot \left[\frac{1}{2} \cdot \left({x}^{3} + 1 - 2 \cdot 3 {x}^{3}\right)\right]}{{x}^{3} + 1} ^ 2$

${y}^{'} = \frac{1}{2} \cdot {x}^{- \frac{1}{2}} \cdot \frac{1 - 5 {x}^{3}}{{x}^{3} + 1} ^ 2$

This can be rewritten as

${y}^{'} = \textcolor{g r e e n}{\frac{1 - 5 {x}^{3}}{2 \sqrt{x} \cdot {\left({x}^{3} + 1\right)}^{2}}}$