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I'm trying to derive the quotient rule in a similar fashion to what was done here: Product rule intuition

I'm trying to get the change in $\frac{f}{g}$.

$$\frac{f + df}{g + dg} - \frac{f}{g} =$$

$$=\frac{gf + gdf} {g * (g + dg)} - \frac{fg + fdg}{g * (g + dg)} =$$

$$=\frac{gdf - fdg}{g * (g + dg)}$$

which is pretty close to the quotient rule. But I can't figure out why the denominator is $g * (g + dg)$ rather than $g^2$. Where am I going wrong?

I'm not interested in deriving the quotient rule in terms of another rule (like the product rule/chain rule). Thanks.

user115350
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Neel
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4 Answers4

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$$f(x+\triangle x)=f(x)+f'(x)\triangle x+O(\triangle x^2)$$ $$g(x+\triangle x)=g(x)+g'(x)\triangle x+O(\triangle x^2)$$ $$h(x)=\frac{f(x)}{g(x)}$$ $$h(x+\triangle x)=h(x)+h'(x)\triangle x+O(\triangle x^2)$$ where $$h'(x)=\frac{f'g-fg'}{g^2}$$ so $$\frac {f(x+\triangle x)}{g(x+\triangle x)} = \frac{f(x)}{g(x)}+\frac{f'g + fg'}{g^2} \triangle x + O(\triangle x^2)$$ and $$\frac {f(x+\triangle x)}{g(x+\triangle x)} \ne \frac{f + f' \triangle x}{g + g' \triangle x} $$

user115350
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Your denominator is $g^2 + g \, dg$, but the second term is negligible compared with $g^2$.

Let $h = f/g$. It may seem more clear if we write your result like this (using infinitesimal intuition): \begin{equation} dh = \left(\frac{g \, \frac{df}{dx} - f \, \frac{dg}{dx}}{g^2 + g \, dg}\right) \, dx. \end{equation} Now the only infinitesimal term in the expression in parentheses is $g \, dg$, so that term is negligible.

Of course, this is just infinitesimal intuition, not a rigorous proof.

littleO
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My proof of quotient rule goes as following: $$\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d}g}{\mathrm{d} z}.\frac{\mathrm{d}z }{\mathrm{d} x}$$ Let: $$g=\frac{1}{f}|z=f$$ So: $$\frac{\mathrm{d} \frac{1}{f}}{\mathrm{d} x}=\frac{\mathrm{d}\frac{1}{f} }{\mathrm{d} f}.\frac{\mathrm{d}f }{\mathrm{d} x}=-\frac{1}{f^{2}}.\frac{\mathrm{d} f}{\mathrm{d} x}=-\frac{f'}{f^{2}}$$ Using the product rule: $$\frac{\mathrm{d} \frac{f}{g}}{\mathrm{d} x}=f'\frac{1}{g}+f(\frac{1}{g})'=\frac{f'}{g}-\frac{fg'}{g^{2}}$$

Felipe A. Barretto
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I don't know if that's formally correct. I will propose another derivation instead. It builds upon the product rule, which you said you have already proved. It also uses the chain rule.

$$\begin{align*} \left(\frac{f(x)}{g(x)}\right)' &= (f(x)g^{-1}(x))' = f'(x)g^{-1}(x) + f(x)(g^{-1}(x))' =\\[0.5ex] &= f'(x)g(x)g^{-2}(x) + f(x)(-g^{-2}(x)g'(x)) =\\[2ex] &= \frac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x)} \end{align*}$$

Note that here $g^{-1}$ does not represent the inverse function of $g$, but its multiplicative inverse.

rubik
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  • Sorry, this is the response that I specifically requested not to receive. – Neel Aug 03 '16 at 17:15
  • @Neel: Ok. I'll leave it here anyway since someone may find it useful. Also, I proposed an alternative because I feel that the way you wanted to derive it was neither formally correct nor intuitive. I say it's unintuitive because what's $\mathrm df$ anyway? A small change in $f$ is not really satisfactory in my opinion. To each their own I guess. – rubik Aug 03 '16 at 17:21
  • https://math.stackexchange.com/a/4261705/945479 – qifeng618 Sep 28 '21 at 01:39