# How do you find the derivative of Arccos(x/2)?

Aug 2, 2016

$\frac{d}{\mathrm{dx}} \left({\cos}^{- 1} \left(\frac{x}{2}\right)\right) = - \frac{1}{\sqrt{4 - {x}^{2}}}$

#### Explanation:

Let's go through the derivation for the general form of the derivative of $\arccos \left(x\right)$

Consider $y = {\cos}^{- 1} \left(x\right)$

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

We can construct a triangle from this. Recall that cosine is adjacent/hypotenuse. This means that adjacent to angle $y$, we have side of length $x$ and that the hypotenuse is of length $1$. By Pythagoras' the length of the opposite side is $\sqrt{1 - {x}^{2}}$.

Now, differentiate our expression:

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

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

But from our triangle we can work out sin(y)!. Sine is opposite/hypotenuse so

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

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

That's the general form, but here we have a function of x inside the function, ie $y \left(u \left(x\right)\right)$. This calls for the chain rule.

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

$u = \frac{x}{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2}$

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

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