# Derivative of inverse function, why do I get this contradiction?

Consider two functions, $f(x)=\sin x$ $\;$ and $g(x)=\arcsin x$. Then, $f'(x)=\cos x$ $\;$ and $g'(x)=\frac{1}{\sqrt{1-x^2}}.$ We know that $g[f(x)]=x$, so $f'(x)=\frac {1}{g'[f(x)]}$ $\;$. $\:$ Substituting in, we have $\cos x=\frac{1}{\sqrt {1-\sin^2 x}}$ $\;$, which yields $\cos x=\frac{1}{\cos x}$ . Why am I getting this result? Thanks!.

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There seems to be a small error in your substitution. – randomwalker Jun 20 '13 at 15:03
Yea I just saw it. – Ovi Jun 20 '13 at 15:04

## 1 Answer

You substituted $1/\sqrt{1-u^2}$ for $1/g'(u)$, when you meant to substitute $1/\sqrt{1-u^2}$ for $g'(u)$.

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Oh ok haha thanks. – Ovi Jun 20 '13 at 15:04