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So I learned that you can use the product rule to guess what the derivative of $\frac{1}{x}$ should be, you just use the fact that $(\frac{1}{x}\cdot x) = 1$ and differentiate both sides and solve for $(\frac{1}{x})'$. My teacher said that there is a similar trick for guessing the derivative of $(\frac{1}{x})$ using the chain rule, I tried using the fact that $\frac{1}{x}\circ\frac{1}{x}= x$ but I am stuck.

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    $\begingroup$ You won’t be able to get it from the fact that $f\circ f= \mathrm{id}$ (what you are trying to do), because $f(x) = \frac{1}{x}$ is not the only function with this property: if you have $g(x) = -x$, then you also have that $g\circ g = \mathrm{id}$, but of course $f’(x)\neq g’(x)$. In short, you are barking up the wrong tree trying to use that property. $\endgroup$ – Arturo Magidin Feb 3 at 23:15
  • $\begingroup$ @ArturoMagidin I think that claim is premature just from your comments. It's the wrong tree if one is going for $f\circ f$, but there could be some other useful compositions. $\endgroup$ – Git Gud Feb 3 at 23:17
  • $\begingroup$ @GitGud: I very carefully did not say anything about using the chain rule in general. I very specifically and explicitly talked about the specific attempt at using the fact that $f\circ f = \mathrm{id}$, which is in fact what the OP is talking about. So, since my comment is only about “going for $f\circ f$,” explicitly (which I specify twice), I have to say your comment is addressing a strawman in which I have no part. $\endgroup$ – Arturo Magidin Feb 3 at 23:20
  • $\begingroup$ @ArturoMagidin I was about to add an adendum to my comment saying that my comment assumed that "property" meant the composition. I stand corrected. $\endgroup$ – Git Gud Feb 3 at 23:22
  • $\begingroup$ Using $x=1/(\frac1x)=u(v(x))$ just leads to $\frac{dv}{du}=\frac{dv}{dx}$, which is obviously true since $u(x) =x. $ So this is the wrong composition to try. $\endgroup$ – timtfj Feb 4 at 1:17
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$$f(x)=\frac1x$$ $$\log \circ \,\,f=-\log x$$ $$(\log \circ \,\,f)’=(-\log x)’$$ $$f’\frac 1f=-\frac1x$$ $$f’=-\frac fx=-\frac 1{x^2}$$

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This works, but I'm not sure if it is allowed. Depends on whether you know the derivative of $\log$.

For any positive real number $x$, set $f(x)=\dfrac 1 x, g(x)=\log(x)$ and $h(x)=-\log(x)$. Note that $g\circ f=h$ and

$$ \begin{align} (g\circ f)(x) = h(x) &\implies g'(f(x))f'(x)=h'(x)\\ &\implies f'(x) = h'(x)/g'(f(x))\\ &\implies f'(x)=-\log'(x)/\log'\left(1/x\right)\\ &\implies f'(x) = -\dfrac{\frac{1}{x}}{\frac{1}{1/x}}\\ &\implies f'(x) = -\dfrac{1}{x^2} \end{align} $$

The crucial point here is that $g'=f^{-1}$. Not unlike what I did here.

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    $\begingroup$ At least you've answered the actual question! $\endgroup$ – timtfj Feb 4 at 1:31

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