# What is the derivative of arcsin(sqrt(1-x^2))?

Aug 14, 2015

$- \frac{x}{\sqrt{{x}^{2} - {x}^{4}}}$

#### Explanation:

Use the Chain Rule and some algebra to simplify:

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

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

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

$= - \frac{x}{\sqrt{{x}^{2}} \sqrt{1 - {x}^{2}}}$

$= - \frac{x}{\sqrt{{x}^{2} - {x}^{4}}}$

This is true for all $x \setminus \in \left(- 1 , 0\right) \cup \left(0 , 1\right)$ (for all $x$ strictly between $- 1$ and $1$ except for $x = 0$).

$y = \arcsin \left(\sqrt{1 - {x}^{2}}\right)$ is actually pretty interesting since you wouldn't expect ahead of time that it would fail to be differentiable at $x = 0$. Here's what the graph of it (red) and its derivative (blue) look like:

The discontinuity in the derivative at $x = 0$ can also be seen by noting that $\sqrt{{x}^{2}} = | x |$ and therefore $- \frac{x}{\sqrt{{x}^{2}}} = - \frac{x}{|} x |$ equals $- 1$ when $x > 0$ and $1$ when $x < 0$. Therefore, $f ' \left(x\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$ when $x > 0$ and $f ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$ when $x < 0$.

For the original function, $f \left(x\right) = \arcsin \left(\sqrt{1 - {x}^{2}}\right)$, evidently it's like taking the graph of $\arcsin \left(x\right)$, shifting it up by $\frac{\pi}{2}$ units, and then reflecting the part of it for $x > 0$ across the horizontal line $y = \frac{\pi}{2}$.

In other words, $\arcsin \left(\sqrt{1 - {x}^{2}}\right) = \arcsin \left(x\right) + \frac{\pi}{2}$ when $x < 0$ and $\arcsin \left(\sqrt{1 - {x}^{2}}\right) = \frac{\pi}{2} - \arcsin \left(x\right)$ when $x > 0$. Weird! Never would have guessed that!

The second equation (for $x > 0$) makes sense if you draw a right triangle, label one angle $\arcsin \left(\sqrt{1 - {x}^{2}}\right)$, label the adjacent side $\sqrt{1 - {x}^{2}}$ and the hypotenuse $1$, then solve for the other side with the Pythagorean Theorem to get $x$. In that situation, $\arcsin \left(x\right)$ would be the complementary angle to the original one, illustrating why $\arcsin \left(\sqrt{1 - {x}^{2}}\right) = \frac{\pi}{2} - \arcsin \left(x\right)$ when $x > 0$.

Here's a graph of the original $f \left(x\right)$ and $f ' \left(x\right)$, along with $\arcsin \left(x\right)$ to illustrate all this: