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"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.

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  • $\begingroup$ "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 $[0,\pi]$. $\endgroup$ – user228113 Dec 9 '17 at 9:50
  • $\begingroup$ 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 $f^{-1}(x)=\text{inverse function}$, $\frac{1}{f(x)}=(f(x))^{-1}=\text{reciprocal of }f(x)$. $\endgroup$ – user228113 Dec 9 '17 at 9:59
  • $\begingroup$ Thanks for your comments. I guess it's my translation which is bad - I'll edit $\endgroup$ – Alex5207 Dec 9 '17 at 10:02
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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, $$(f^{-1})’(\frac{1}{\sqrt{2}})=\frac{1}{f’(\frac{\pi}{4})}=\frac{1}{-\sin \frac{\pi}{4}}=-\frac{1}{\frac{1}{\sqrt{2}}}=-\sqrt2$$

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  • $\begingroup$ Hi Rohan. Not quite sure i understand the notation with the square brackets - Could you clarify? $\endgroup$ – Alex5207 Dec 9 '17 at 9:31
  • $\begingroup$ @Rohan You've missed out the differential operator. $\endgroup$ – Botond Dec 9 '17 at 9:34
  • $\begingroup$ Corrected. Thanks for pointing out the typo @Botond. $\endgroup$ – Rohan Dec 9 '17 at 9:36
  • $\begingroup$ @Rohan. Could you go through the example with $f(x)=cos(x)$? I think that might help in understanding it $\endgroup$ – Alex5207 Dec 9 '17 at 9:43
  • $\begingroup$ +1 But it is not a definition, it's a theorem $\endgroup$ – Ant Dec 9 '17 at 13:38
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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.

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

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