Reciprocal derivative of function: $(f^{-1})'(\frac{1}{2} \sqrt{2})$ for $f(x)=\cos x$

Question

"The function $f(x)=\cos(x)$ has an inverse/reciprocal (Not sure which one is the correct translation) function arccos in the interval $[0;\pi]$ Determine the derivative of $$(f^{-1})'(\frac{1}{2} \sqrt{2})$$

So, I guess the steps to take are

1. Find the inverse of $\cos(x)$ which is $\frac{1}{\cos(x)}$
2. Find the derivative of that function which I belive is $(\frac{1}{\cos(x)})'=\frac{\sin(x)}{\cos^2(x)}$
3. Insert the value: $\frac{\sin(\frac{1}{2}\sqrt{2})}{\cos^2(\frac{1}{2}\sqrt{2})}$$=\frac{\frac{\pi}{4}}{\frac{\pi^2}{4^2}}$$=\frac{\pi}{4} \cdot \frac{16}{\pi^2}$$=\frac{4}{\pi}$

However, that answer is not correct. I'm told that the correct answer is $-2^{\frac{1}{2}}$ which I don't understand.

Answer 1

Use the definition $$(f^{-1})’(x)=\frac{1}{f’(f^{-1}(x))}$$ and see what you get. Note that $f(x)=\cos x$ and $f’(x)=-\sin x$. Also, that $f^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$.

EDIT:

We need $(f^{-1})’(\frac{1}{\sqrt{2}})$. Using the definition, we get, $$(f^{-1})’(\frac{1}{\sqrt{2}})=\frac{1}{f’(\frac{\pi}{4})}=\frac{1}{-\sin \frac{\pi}{4}}=-\frac{1}{\frac{1}{\sqrt{2}}}=-\sqrt2$$

Answer 2

The problem here is to figure out if they mean functional inverse or multiplicative inverse. Arccos is the functional inverse. The function which takes us back to $x$ if we plug in $\cos(x)$ into it.

So they are probably asking for functional inverse and not multiplicative inverse.

Answer 3

At a particular point say, [ x, f(x) ] , Derivative of (f inverse (x)) = 1/f'(x) . At (f inverse (1/√2)) where f(x) = cos(x), x= π/4 since 'x' lies between (0,π). Derivative of cos(x) is -sin(x). Put x=π/4. Hence, derivative of (f inverse x) is 1/(-1/√2) = - √2