Using the chain rule to guess the derivative of $\frac{1}{x}$

Question

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.

Answer 1

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.

Answer 2

$$f(x)=\frac1x$$ $$\log \circ \,\,f=-\log x$$ $$(\log \circ \,\,f)’=(-\log x)’$$ $$f’\frac 1f=-\frac1x$$ $$f’=-\frac fx=-\frac 1{x^2}$$