How to show if two derivatives are equivalent?

Question

I have completed parts a and b to the attached problem, but I am having a hard time with part c. Can anyone tell me I'm going in the right direction or if I am missing something?

Answer 1

They just want you to show using the quotient rule to differentiate $$f(x)=\frac{k}{g(x)}$$ yields the same answer as using the product rule to differentiate $$f(x)=k\left(g(x)\right)^{-1}$$

For the quotient rule: $\left(\frac{k}{g}\right)'=\frac{gk'-g'k}{g^2}$ we have $$f'(x)=\frac{g(x)\left(k\right)'-kg'(x)}{g^2(x)}=\frac{0-kg'(x)}{g^2(x)}=-\frac{kg'(x)}{g^2(x)}$$

and using the product rule $(kg)'=k'g+g'k$ we have $$f'(x)=(k)'(g(x))^{-1}+k(-1)(g(x))^{-2}\cdot g'(x)=\frac{k'}{g(x)}+\frac{-k}{g^2(x)}\cdot g'(x)=-\frac{kg'(x)}{g^2(x)}$$