# How do you differentiate  g(x) = sqrtarccos(x^2-1) ?

Mar 2, 2016

Use implicit differentiation and the chain rule to find that

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

#### Explanation:

First, let's see what the derivative of the $\arccos$ function is by using implicit differentiation:

Let $y = \arccos \left(x\right)$

$\implies \cos \left(y\right) = x$

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

$\implies - \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left(y\right)$

$= - \frac{1}{\sin} \left(\arccos \left(x\right)\right)$

$= - \frac{1}{\sqrt{1 - {x}^{2}}}$
(For the last step, try drawing a right triangle where $\cos \left(\theta\right) = x$ and then see what $\sin \left(\theta\right)$ is equal to)

With that, the remainder of the problem can be done using the chain rule:

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

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

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

=(2x)/(2sqrt(arccos(x^2-1))sqrt((1+x^2-1)(1-x^2+1))#

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