Alternate form of quotient rule

Question

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?

Answer 1

Well, think about it.

$g(x)$ 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): $$ \left(g_1(x)=\dfrac{f(x)}{h(x)}=\dfrac{\sin(x)}{\sinh(x)}=\dfrac{e^{ix}-e^{-ix}}{e^x-e^{-x}} \right) $$

I cannot comment on the visual appeal of your geometric understanding of it as i don't really have that, but that's also subjective.