# How do you find the derivative of the function: (arccos(x/6))^2?

Jan 23, 2017

Let $h \left(x\right) = \frac{x}{6}$
Let $g \left(h\right) = \arccos \left(h\right)$
Let $f \left(g\right) = {g}^{2}$
To differentiate, use the chain rule with a function nested within a function within a function.

#### Explanation:

The chain rule with a function nested within a function within a function:

$\frac{d \left(\left(\arccos {\left(\frac{x}{6}\right)}^{2}\right)\right)}{\mathrm{dx}} = \frac{d \left(f \left(g \left(h \left(x\right)\right)\right)\right)}{\mathrm{dx}}$

$\frac{d \left(\left(\arccos {\left(\frac{x}{6}\right)}^{2}\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \frac{\mathrm{dg}}{\mathrm{dh}} \frac{\mathrm{dh}}{\mathrm{dx}} \text{ [1]}$

$\frac{\mathrm{df}}{\mathrm{dg}} = 2 g = 2 \arccos \left(h\right) = 2 \arccos \left(\frac{x}{6}\right)$

$\frac{\mathrm{dg}}{\mathrm{dh}} = - \frac{1}{\sqrt{1 - {h}^{2}}} = - \frac{1}{\sqrt{1 - {\left(\frac{x}{6}\right)}^{2}}}$

$\frac{\mathrm{dh}}{\mathrm{dx}} = \frac{1}{6}$

Substituting into equation [1]

$\frac{d \left(\left(\arccos {\left(\frac{x}{6}\right)}^{2}\right)\right)}{\mathrm{dx}} = \left(2 \arccos \left(\frac{x}{6}\right)\right) \left(- \frac{1}{\sqrt{1 - {\left(\frac{x}{6}\right)}^{2}}}\right) \left(\frac{1}{6}\right)$

$\frac{d \left(\left(\arccos {\left(\frac{x}{6}\right)}^{2}\right)\right)}{\mathrm{dx}} = \frac{- 2 \arccos \left(\frac{x}{6}\right)}{6 \sqrt{1 - {\left(\frac{x}{6}\right)}^{2}}}$

$\frac{d \left(\left(\arccos {\left(\frac{x}{6}\right)}^{2}\right)\right)}{\mathrm{dx}} = \frac{- 2 \arccos \left(\frac{x}{6}\right)}{\sqrt{36 - {x}^{2}}}$