Prove $\operatorname{arccosh}'(y) = \frac{1}{\sqrt{y^2 - 1}}$

Question

I have to prove that Prove $\operatorname{arccosh}'(y) = \cfrac{1}{\sqrt{y^2 - 1}}$ for all $y \in (1, \infty)$

I have to do this using the formula for differentiation of inverse functions.

$$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$$

For some reason the answer i end up with is

$$\operatorname{arccosh}'(y) = \cfrac{1}{y + \sqrt{y^2 - 1}}.$$

Which is obviously wrong. So how should i go about solving this?

Answer 1

Let $f = \cosh$. Then $$ (f^{-1})'(y) = \frac{1}{f' \circ f^{-1}(y)} = \frac{1}{\sinh \circ \cosh^{-1} y} $$ You know that $(\cosh)^2 - (\sinh)^2 = 1$. Hence $$\sinh \circ \cosh^{-1} y = \sqrt{(\cosh \circ \cosh^{-1} y)^2 - 1} = \sqrt{y^2 - 1} $$

Answer 2

$$\operatorname{argcosh}'(y)=\frac1{\sinh(\operatorname{argcosh}y)}$$ Now $\cosh(\operatorname{argcosh}y)=y$, and $\cosh^2y-\sinh^2y=1$. Can you continue?