Derivative of Arcsine Function

From ProofWiki
Jump to navigation Jump to search

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

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


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:

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

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Derivative_of_Arcsine_Function&oldid=625638"