How do you differentiate  f(x)= xsqrt(1-x^2) using the product rule?

Dec 25, 2015

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

Explanation:

The product rule: $\frac{d}{\mathrm{dx}} \left[g \left(x\right) h \left(x\right)\right] = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

$\textcolor{w h i t e}{\times} f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[x\right] \cdot \sqrt{1 - {x}^{2}} + x \frac{d}{\mathrm{dx}} \left[\sqrt{1 - {x}^{2}}\right]$

Find each derivative separately.

$\textcolor{w h i t e}{\times} \frac{d}{\mathrm{dx}} \left[x\right] = 1$

Use the chain rule—recall that $\frac{d}{\mathrm{dx}} \left[{u}^{\frac{1}{2}}\right] = \frac{1}{2} {u}^{- \frac{1}{2}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

$\textcolor{w h i t e}{\times} \frac{d}{\mathrm{dx}} \left[{\left(1 - {x}^{2}\right)}^{\frac{1}{2}}\right] = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left[- {x}^{2}\right]$

$\textcolor{w h i t e}{\times} \implies \frac{1}{2 \sqrt{1 - {x}^{2}}} \left(- 2 x\right) = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Plug these back in:

$\textcolor{w h i t e}{\times} f ' \left(x\right) = \sqrt{1 - {x}^{2}} - {x}^{2} / \sqrt{1 - {x}^{2}}$

$\textcolor{w h i t e}{\times} \implies \frac{1 - 2 {x}^{2}}{\sqrt{1 - {x}^{2}}}$