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

"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}$$=\frac{4}{\pi} However, that answer is not correct. I'm told that the correct answer is -2^{\frac{1}{2}} which I don't understand. • "Reciprocal of f" = "\frac1f". "Inverse of f" = (if f is bijective) "the only map g such that f(g(y))=y for all y in the codomain and x=g(f(x)) for all x in the domain". \arccos is the inverse of the restriction of \cos to the interval [0,\pi]. – user228113 Dec 9 '17 at 9:50 • There is an inherent ambiguity in using x^{-1} for the reciprocal of a quantity and f^{-1} for the inverse of a function. Since both notations are widely used, it is standard to assume that context is paramount: (\text{polynomial or something explicit})^{-1}=\frac{1}{\text{polynomial or something explicit}}=\text{reciprocal}; (\text{function})^{-1}=\text{inverse} and \frac1{\text{function}}=\text{reciprocal}. For instance f^{-1}(x)=\text{inverse function}, \frac{1}{f(x)}=(f(x))^{-1}=\text{reciprocal of }f(x). – user228113 Dec 9 '17 at 9:59 • Thanks for your comments. I guess it's my translation which is bad - I'll edit – Alex5207 Dec 9 '17 at 10:02 ## 3 Answers Use the definition$$(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$$• Hi Rohan. Not quite sure i understand the notation with the square brackets - Could you clarify? – Alex5207 Dec 9 '17 at 9:31 • @Rohan You've missed out the differential operator. – Botond Dec 9 '17 at 9:34 • Corrected. Thanks for pointing out the typo @Botond. – Rohan Dec 9 '17 at 9:36 • @Rohan. Could you go through the example with$f(x)=cos(x)$? I think that might help in understanding it – Alex5207 Dec 9 '17 at 9:43 • +1 But it is not a definition, it's a theorem – Ant Dec 9 '17 at 13:38 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.

• Thanks for answering. As you say, it's the functional inverse that we're asked for – Alex5207 Dec 9 '17 at 9:29

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