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

Dec 10, 2016

We use the difference rule to differentiate the entire relation. However, let's differentiate $\arcsin x$ and $\sqrt{1 - {x}^{2}}$ individually.

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$y = \arcsin x \to \sin y = x$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

Using the identity ${\cos}^{2} y = \sqrt{1 - {\sin}^{2} y}$:

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

Since $\sin y = x$, ${\sin}^{2} y = {x}^{2}$.

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

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

We let $y = \sqrt{u} = {u}^{\frac{1}{2}}$ and $u = 1 - {x}^{2}$. Differentiating each, we get $y ' = \frac{1}{2 {u}^{\frac{1}{2}}}$ and $u = - 2 x$.

By the chain rule:

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

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

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

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

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We can now combine the two derivatives.

Call the initial function $f ' \left(x\right)$:

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

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

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

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

f'(x) = (2(1 + x))/(2sqrt(1 - x^2)

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

You may want to rationalize the denominator, depending on your teacher's wishes.

Hopefully this helps!