Using the chain rule to solve reciprical of a function

Question

Here is my issue.

To prove the quotient rule from the product rule, we require the use of chain rule on $(h(x))^{-1}$. Why?

For example, if $h(x)=3x^{-1}$, then $h'(x)=-3x^{-2}$

However, the chain rule says $\frac{d}{dx} (h(x))^{-1} = -1(h(x))^{-2} h'(x)$, which would suggest for $h(x)=3x^{-1}$, then $h'(x)=-3(3x^{-2})$

Answer 1

In your solution, $\frac{d}{dx}h(x)^{-1}$ i.e. $h(x) = \frac{1}{3}x$ not $3x^{-1}$

Answer 2

I can give an example let $f(x)=\frac{1}{x}$ so by product rule $f'(x)=1'x^{-1}+1.(x^{-1})'=\frac{-1}{x^2}$ and by normal derivative rule $f'(x)=nx^{n-1}$ we get the same result. And chain rule we get the same result. In your example f'(x) for $3x^{-1}$ is $-3x^{-1-1}.3x'$ is correct. In chain rule you didnt differentiate $3x$ so you are getting different results.so $f'(3x^{-1})=-1.3x^{-2}.(3x)'=-3.(3x^{-2})$