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

Jul 29, 2014

$y ' = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Using Chain Rule

$y = \sqrt{f \left(x\right)}$

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

Similarly, following for the above function,

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

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

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

$y ' = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Jun 3, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

Another method is using implicit differentiation like this:

$y = \sqrt{1 - {x}^{2}}$

${y}^{2} = 1 - {x}^{2}$

${x}^{2} + {y}^{2} = 1$

Now differentiate with respect to $x$:

$2 x + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$x + y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - x$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \frac{x}{y}$

Finally, make the substitution $y = \sqrt{1 - {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Jun 3, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

$y = \sqrt{1 - {x}^{2}}$

Now let $u = 1 - {x}^{2}$

$y = {u}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2} \left({u}^{- \frac{1}{2}}\right) = \frac{1}{2 {u}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{u}} = \frac{1}{2 \sqrt{1 - {x}^{2}}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$

Now apply the chain rule:

$\frac{\mathrm{dy}}{\cancel{\mathrm{du}}} \times \frac{\cancel{\mathrm{du}}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \times \frac{1}{2 \sqrt{1 - {x}^{2}}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$