# How do you find the derivative of arccos ([2x + 1]/2)?

May 5, 2015

$\frac{d}{\mathrm{dx}} \arccos \left(\frac{2 x + 1}{2}\right) = \frac{- 1}{\setminus \sqrt{\frac{3}{4} - x - {x}^{2}}}$

First I will prove a result needed to do the differentiation:

$\frac{d}{\mathrm{du}} \arccos \left(u\right) = \frac{- 1}{\setminus \sqrt{1 - {u}^{2}}}$

Proof:
Let $y = \arccos \left(u\right)$, take the cosine of each side.

$\setminus \implies \cos \left(y\right) = u$, now differentiate each side with respect to $u$

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

$\setminus \implies \frac{\mathrm{dy}}{\mathrm{du}} = \frac{d}{\mathrm{du}} \left[\arccos u\right] = \frac{- 1}{\sin \left(y\right)}$

Draw a triangle that agrees with the earlier expression, $\cos \left(y\right) = u$ to find out what $\sin \left(y\right)$ is equal to. From the diagram below, $\sin \left(y\right) = \setminus \sqrt{1 - {u}^{2}}$

Substitute $\sin \left(y\right)$ to get the desired result:$\frac{d}{\mathrm{du}} \arccos \left(u\right) = \frac{- 1}{\setminus \sqrt{1 - {u}^{2}}}$

Now use the chain rule to take the derivative of $y = \arccos \left(\frac{2 x + 1}{2}\right)$

Let $u = \frac{2 x + 1}{2}$ so that $y = \arccos \left(u\right)$

Chain Rule
$\frac{d}{\mathrm{dx}} y = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \left(\frac{- 1}{\setminus \sqrt{1 - {u}^{2}}}\right) \left(1\right)$

Substitute $u$ in the above result to get

$\frac{d}{\mathrm{dx}} \arccos \left(\frac{2 x + 1}{2}\right) = \frac{- 1}{\setminus \sqrt{\frac{3}{4} - x - {x}^{2}}}$