Derivative of Arcsine Function

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Theorem

Let $x \in \R$ be a real number such that $\size x < 1$, that is, $\size {\arcsin x} < \dfrac \pi 2$.

Let $\arcsin x$ be the real arcsine of $x$.


Then:

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


Corollary

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


Proof

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

Then:

$x = \sin y$

Then from Derivative of Sine Function:

$\dfrac {\d x} {\d y} = \cos y$

Hence from Derivative of Inverse Function:

$\dfrac {\d y} {\d x} = \dfrac 1 {\cos y}$

From Sum of Squares of Sine and Cosine, we have:

$\cos^2 y + \sin^2 y = 1 \implies \cos y = \pm \sqrt {1 - \sin^2 y}$

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:

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

$\blacksquare$


Also see


Sources