# How do you find the derivative of g(x)= x^2sqrt(1-x^2)?

Aug 19, 2015

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

#### Explanation:

You can use the product rule and the chain rule to differentiate this function, which can be rewritten as

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

The product rule allows you to differentiate functions that take the form

$\textcolor{b l u e}{g \left(x\right) = f \left(x\right) \cdot h \left(x\right)}$

by using the formula

color(blue)(d/dx(g(x)) = [d/dx(f(x))] * h(x) + f(x) * d/dx(h(x))

$f \left(x\right) = {x}^{2} \text{ }$ and $\text{ } h \left(x\right) = {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

This means that you can write

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

To find $\frac{d}{\mathrm{dx}} \left(h \left(x\right)\right)$, you need to use the chain rule for ${u}^{\frac{1}{2}}$, with $u = \left(1 - {x}^{2}\right)$. This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{d}{\mathrm{du}} {u}^{\frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} \cdot {u}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)$

$\frac{d}{\mathrm{dx}} {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x\right)$

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

Plug this into your target derivative to get

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

This can be simplified to get

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

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

Finally, you can rewrite this as

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