# How do you find the derivative of arccos(x^2)?

Feb 20, 2015

The answer is: $y ' = - \frac{2 x}{\sqrt{1 - {x}^{4}}}$

Since y=arccosf(x)rArry'=-f(x)/sqrt(1-[f(x)]^2,

than:

$y ' = - \frac{2 x}{\sqrt{1 - {x}^{4}}}$.

Feb 21, 2015

We have

$y = \arccos \left({x}^{2}\right)$

Take the cosine of both sides

$\cos \left(y\right) = \cos \left(\arccos \left({x}^{2}\right)\right)$

$\cos \left(y\right) = {x}^{2}$

Now differentiate both sides with respect to $x$

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

Divide both sides by $- \sin \left(y\right)$

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

Recall that ${\sin}^{2} y + {\cos}^{2} y = 1$ Therefore,

${\sin}^{2} y = 1 - {\cos}^{2} y$

$\sin \left(y\right) = \sqrt{1 - {\cos}^{2} y}$ we restrict ourselves to the positive root

From above $\cos \left(y\right) = {x}^{2}$. Consequently

${\cos}^{2} y = {x}^{4}$

So we can write

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