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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.

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  • $\begingroup$ What is the 'function theorem'? I think its correct name is 'inverse function theorem'. $\endgroup$
    – Hanul Jeon
    Commented Nov 19, 2014 at 3:19

1 Answer 1

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

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

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  • $\begingroup$ for the third step, are you using the trigonometric identity $\cos^{2}y + sin^{2}y = 1$? $\endgroup$
    – Mahidevran
    Commented Nov 19, 2014 at 3:35
  • $\begingroup$ Yes, it is. And notice that $\cos y >0$ when $y\in (-\frac{\pi}{2}, \frac{\pi}{2})$. $\endgroup$
    – Paul
    Commented Nov 19, 2014 at 3:40
  • $\begingroup$ @Mahidevran: Could you upvote me so that I will get 10k reputation:) $\endgroup$
    – Paul
    Commented Nov 19, 2014 at 3:44
  • $\begingroup$ I accepted your answer as the official answer, and I pressed the arrow that goes up. It turned red and the check mark green. I hope that does it. Thanks:) $\endgroup$
    – Mahidevran
    Commented Nov 19, 2014 at 3:48
  • $\begingroup$ @Mahidevran: Thanks. Just a joke. May you happy! $\endgroup$
    – Paul
    Commented Nov 19, 2014 at 3:49

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