Derivative of Arcsine Function/Proof
Jump to navigation
Jump to search
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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Inverse Ratios
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.5 \ (3)$