# Correct simultanous application of chain and product rule

For two continuous differentiable functions $g(x)$ and $h(x)$ we seek

$$\frac{d}{dx} [g(x) h^{-1}(x)]$$

where $h^{-1}(x) = \frac{1}{h(x)}$. This asks us to apply product and chain rule in sequence, I believe. Is the following correct?

$$\frac{d}{dx} [g(x) h^{-1}(x)] = g'(x)h^{-1}(x) + g(x) [-h^{-2}(x)h'(x)] = g'(x)h^{-1}(x) - g(x) h'(x) h^{-2}(x)$$

I am not sure about the application of the chain rule in the second term (first equation), in particular.

• $h^{-1}(x)$ usually denotes the inverse function of $h$, not $\frac{1}{h(x)}$ (which would be denoted $(h(x))^{-1}$). It looks like you're trying to differentiate $\frac{g(x)}{h(x)}$, which is just the quotient rule; are you trying to derive that? – DylanSp Feb 2 '16 at 16:05
• yes. I will search for quotient rule now. – tomka Feb 2 '16 at 16:06

## 1 Answer

Your reasoning is fully correct, however, you should not use $f^{-1}(x)$ to denote $\frac{1}{f(x)}$, as by convention it is used to denote the inverse of a function $f^{-1}(f(x))=x$. Write $\frac{1}{f(x)}$ or $(f(x))^{-1}$ for the reciprocal of a function instead.

As for the task, you could simply use the quotient rule, which gives exactly what you are after: $$\frac{d}{dx} \left[ \frac{g(x)}{h(x)}\right] = \frac{h(x)g'(x)-g(x)h'(x)}{(h(x))^2}=g'(x)(h(x))^{-1}-g(x)h'(x)(h(x))^{-2}$$

However, your task was to use the product rule and chain rule only, I assume. Notice that you get exactly the same answer (so yes, your answer is correct), meaning that you have actually proved the quotient rule by using the chain and product rule only.

• Thank you - that's encouraging and a complete answer. – tomka Feb 2 '16 at 16:24