Derivative of Arcsine Function/Proof

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Theorem

$\dfrac {\map \d {\arcsin x} } {\d x} = \dfrac 1 {\sqrt {1 - x^2} }$


Proof

Let $y = \arcsin x$ where $-1 < x < 1$.


Then:

\(\ds x\) \(=\) \(\ds \sin y\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d x} {\d y}\) \(=\) \(\ds \cos y\) Derivative of Sine Function


Then:

\(\ds \cos^2 y + \sin^2 y\) \(=\) \(\ds 1\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds \cos y\) \(=\) \(\ds \pm \sqrt {1 - \sin^2 y}\)


Now $\cos y \ge 0$ on the image of $\arcsin x$, that is:

$y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}$

Thus it follows that we need to take the positive root of $\sqrt {1 - \sin^2 y}$.


So:

\(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac 1 {\sqrt {1 - \sin^2 y} }\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {1 - x^2} }\)

$\blacksquare$


Sources