# Alternate form of quotient rule

Given a function $$f(x)=h(x)g(x)$$, the product rule gives us $$\begin{equation} f'(x)=g(x)h'(x)+h(x)g'(x) \tag{1} \end{equation}$$

We can rearrange this to get the quotient rule, as follows.

Given a function $$g(x)=\frac{f(x)}{h(x)}, \; h(x) \neq 0$$, equation 1 gives us \begin{align} g'(x)&=\frac{f'(x)-g(x)h'(x)}{h(x)}\\ &=\frac{f'(x)h(x)-g(x)h(x)h'(x)}{[h(x)]^2}\\ &=\frac{f'(x)h(x)-f(x)h'(x)}{[h(x)]^2} \tag{2} \end{align}

I can see the standard version of the quotient rule in line 3 of equation 2 has the advantage(?) of not mentioning $$g$$ on the RHS, but I find line 1 easier to understand geometrically. Is there any reason not to use line 1 as a version of the quotient rule?

• My preferred form is $$\frac{g'(x)}{g(x)}=\frac{f'(x)}{f(x)}-\frac{h'(x)}{h(x)}.$$ Yes I know this is problematical where $f$ vanishes.... Dec 7, 2019 at 19:37
• I can see it looks neat algebraically, but if anything it's less intuitive to me geometrically.
– mjc
Dec 7, 2019 at 19:46

$$g(x)$$ is not a function you know explicitly, if $$f(x)$$ and $$h(x)$$ are different 'classes' of functions you will have a hard time figuring out $$h(x)$$, if both of them are polynomials then you still have to divide them, which takes up a considerable amount of time depending on your division skills and the length of the polynomials.
• Easy case: $$\left(g_1(x)=\dfrac{f(x)}{h(x)}=\dfrac{\sin(x)}{\cos(x)}\right)'\Rightarrow g_1(x)=\tan(x)$$
• Harder case: $$\left(g_2(x)=\dfrac{f(x)}{h(x)}=\dfrac{x^7+2x^6+2x^4-x^2}{x^3-5x+7}\right)'\Rightarrow g_2(x)=x4+x3+x2+x+1 \text{ with a remainder of } -3x^2-2x-7$$ Of course you could just calculate with $$\frac{f(x)}{h(x)}$$ but at that point you are using the classical quotient rule.
• Non-solvable case (by simple algebraic manipulations): $$\left(g_1(x)=\dfrac{f(x)}{h(x)}=\dfrac{\sin(x)}{\sinh(x)}=\dfrac{e^{ix}-e^{-ix}}{e^x-e^{-x}} \right)$$