How do you differentiate y = x * sqrt (4 - x^2)?

Jan 19, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(2 - {x}^{2}\right)}{\sqrt{4 - {x}^{2}}}$

Explanation:

differentiate using the 'product rule' and 'chain rule':

$\frac{\mathrm{dy}}{\mathrm{dx}} = x . \frac{d}{\mathrm{dx}} {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} . \frac{d}{\mathrm{dx}} \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(\frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} . \frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right)\right) + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} . 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(\frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} . \left(- 2 x\right)\right) + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$

[ common factor of  (4 - x^2 )^(-1/2) ]

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \left[- {x}^{2} + 4 - {x}^{2}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} . \left(4 - 2 {x}^{2}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(2 - {x}^{2}\right)}{\sqrt{4 - {x}^{2}}}$