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In calculus class, as a shortcut for $\dfrac{\mathrm d}{\mathrm dx} \dfrac{f(x)}{g(x)}$ I'd often just do $\dfrac{\mathrm d}{\mathrm dx} f(x) g(x)^{-1}$ and then use the multiplication rule. I'd do the same with integration.

Is there any harm with this approach? I don't know of ant special case where perhaps 0 might mess up the equation.

Lord_Farin
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jaysonpowers
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  • implicit in the calculation of the derivative is that $g(x)$ is nonzero. it's fine to do it that way if you can't remember the quotient rule (i only can because i've had to mark hundreds of calculus assignments and exams). it's slightly ambiguous as to what you mean by "I'd do the same with integration" though – citedcorpse Jun 25 '13 at 18:04

2 Answers2

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You could use logarithms instead. let $$y=\frac{f(x)}{g(x)}$$ $$\ln y=\ln f(x)-\ln g(x)$$ $$(\frac1y)(\frac{dy}{dx})=\frac{1}{f(x)}\frac{d}{dx}f(x)-\frac{1}{g(x)} \frac{d}{dx}g(x)$$ $$\frac{dy}{dx}=y\left(\frac{1}{f(x)}\frac{d}{dx}f(x)-\frac{1}{g(x)} \frac{d}{dx}g(x)\right)$$ $$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f(x)}{g(x)}\left(\frac{1}{f(x)}\frac{d}{dx}f(x)-\frac{1}{g(x)} \frac{d}{dx}g(x)\right)$$

user80551
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You can also do this by an easy implicit differentiation and avoid having to know how to differentiate $\left(g(x)\right)^{-1},$ but the price to pay for this is a little extra algebra:

$$ y \; = \; \frac{f}{g}$$ $$ y \cdot g \; = \; f$$ $$ \left(y \cdot g \right)' \; = \; f' $$ $$ y'g + yg' \; = \; f' $$ $$ y'g \; = \; f' - yg' $$ $$ y' \; = \; \frac{f' - yg'}{g} $$ $$ y' \; = \; \frac{f' - \left(\frac{f}{g}\right)g'}{g} $$ $$ y' \; = \; \frac{f' - \left(\frac{f}{g}\right)g'}{g} \cdot \frac{g}{g}$$ $$ y' \; = \; \frac{f'g - fg'}{g^2} $$

Dave L. Renfro
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