# How do you differentiate f(x)= x/sqrt(1-x^2)  using the quotient rule?

Dec 30, 2015

We can see this as a product: $f \left(x\right) = x \cdot {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$ and use the product rule, as well as - for the second term - the chain rule.

#### Explanation:

• Chain rule states that for $y = f \left(x\right) g \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$
• Chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Renaming $u = 1 - {x}^{2}$, we get that the second term becomes ${u}^{- \frac{1}{2}}$ and chain rule turns out to be applicable.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \cdot {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} + x \left(\frac{x}{{u}^{\frac{3}{2}}}\right)$

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

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