Derivative of $\sec^{-1}(\frac{x}{3})$
I have tried these types of problems with two different approaches and keep getting the same answer which seems to be wrong. I suspect I am doing something obvious incorrectly; however, I can't seem to figure it out.
First method:
$y = \sec^{-1}(\frac{x}{3})$
therefore,
$\sec(y) = \frac{x}{3}$
$\frac{dy}{dx}\sec(y)\tan(y) = \frac{1}{3}$
$\frac{dy}{dx} = \frac{1}{3\sec(y)\tan(y)}$
Since $\sec(y) = \frac{x}{3}$ and $\tan(y) = \sqrt{\sec^2(y) -1} = \sqrt{\frac{x^2}{9} - 1}$
$\frac{dy}{dx} = \frac{1}{x\sqrt{\frac{x^2}{9} - 1}}$
Method two:
$(f')^{-1}(\frac{x}{3}) = \frac{1}{f'(\sec^{-1}(\frac{x}{3}))}$
$ = \frac{1}{3\sec(\sec^{-1}{(\frac{x}{3})})\tan(\sec^{-1}(\frac{x}{3}))}$
$= \frac{1}{x\sqrt{\sec^2(\sec^{-1}(\frac{x}{3})) - 1}}$
$= \frac{1}{x\sqrt{\frac{x^2}{9} -1}}$