# A calculus of the absurd

##### 10.10.5 Derivative of $$\arcsin (x)$$

This proof/derivation is included purely for completeness and is mechanistically almost entirely the same as the cases of $$\arctan$$ (Section 10.10.3) and $$\arccos$$ (Section 10.10.4).

We start with the definition of $$\arcsin$$, which is that

$$\arcsin (\sin (x)) = x.$$

Differentiating this,

\begin{align} & \frac {d}{dx} \left (\arcsin (\sin (x))\right ) = \frac {d}{dx} (1) \\ & \iff \frac {d}{d(\sin (x))} (\arcsin (\sin (x))) \frac {d}{dx} \sin (x) = 1 \\ & \iff \frac {d}{d(\sin (x))} (\arcsin (\sin (x))) \cos (x) = 1 \\ & \iff \frac {d}{d(\sin (x))} (\arcsin (\sin (x))) = \frac {1}{\cos (x)} \end{align}

Then, we can substitute $$u = \sin (x)$$, using which

$$\frac {d}{du}\arcsin (u) = \frac {1}{\cos (x)}.$$

This is almost what we would like, except that we have a random $$\cos (x)$$ floating around. This can be removed by noting that $$\cos (x) = \sqrt {1 - \sin ^2(x)}$$, or $$\cos (x) = \sqrt {1 - u^2}$$, i.e.

$$\frac {d}{du}\arcsin (u) = \frac {1}{\sqrt {1 - u^2}}$$