Using the Inverse Function Theorem prove that $(\sin^{-1}x)'$ = $\frac{1}{\sqrt{1-x^2}}$.

Question

Using the Inverse Function Theorem prove that $(\sin^{-1}x)'$ = $\frac{1}{\sqrt{1-x^2}}$.

Proof: Let $f(x) = \sin x$, for $x$ in $(-1,1)$. Then let $x_{0}$ be in (-1,1).

Then $f'(x_{0})$ = $\cos(x_{0})\neq 0$, where $f'(x_{0})$ $> 0$. So $f$ is increasing thus one to one.

Then using the Inverse Function Theorem, $(f^{-1})'(y_{0})$ = $\frac{1}{\cos(\sin^{-1}y_{0})}$ = $sec(x_{0})$ = $\frac{1}{\sqrt{1-y_{0}^2}}$ .

Is this a a correct way to prove it? Please any suggestion/feedback would be appreciated. And can someone please verify the intervals are correct since I am working with inverses. Thank you very much.

Answer 1

Let $y=\sin^{-1}x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. So $x=\sin y$.

$$(\sin^{-1}x)'=\frac{1}{\sin' y} = \frac{1}{\cos y}= \frac{1}{\sqrt{1-\sin^2y}} = \frac{1}{\sqrt{1-x^2}}$$