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

\begin{equation} \arcsin (\sin (x)) = x. \end{equation}

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

\begin{equation} \frac {d}{du}\arcsin (u) = \frac {1}{\cos (x)}. \end{equation}

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.

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