# How do you find the derivative of the function: sin[arccos(x)]?

Oct 21, 2016

d/(dx) sin(arccos(x)) = -x/(sqrt(1-x^2)

#### Explanation:

$f \left(x\right) = \sin \left(\arccos \left(x\right)\right)$

Assuming that we want to consider this as a Real valued function of Real numbers, the domain of $f \left(x\right)$ is $\left[- 1 , 1\right]$

When $x \in \left[- 1 , 1\right]$ then $\arccos \left(x\right) \in \left[0 , \pi\right]$ and hence:

$\sin \left(\arccos \left(x\right)\right) \ge 0$

Then from Pythagoras, we have:

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

Hence:

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

using the non-negative square root since we have already established that $\sin \left(\arccos \left(x\right)\right) \ge 0$

So:

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

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sin \left(\arccos \left(x\right)\right)} = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right)$

color(white)(d/(dx) sin(arccos(x))) = -x/(sqrt(1-x^2)