# 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

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