# How do you use the quotient rule to differentiate x / (1-x^2)^(1/2)?

##### 1 Answer
Apr 8, 2017

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

#### Explanation:

Quotient rule states if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$

Here in $\frac{x}{1 - {x}^{2}} ^ \left(\frac{1}{2}\right)$, we have $g \left(x\right) = x$ and $\therefore \frac{\mathrm{dg}}{\mathrm{dx}} = 1$

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

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

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

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

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

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

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