Calculus/Quotient Rule

Quotient rule

There rule similar to the product rule for quotients. To prove it, we go to the definition of the derivative:


 * $$\begin{align}

\frac{d}{dx} \frac{f(x)}{g(x)} &= \lim_{h \to 0} \frac{\frac{f(x + h)}{g(x + h)} - \frac{f(x)}{g(x)}}{h} \\ &= \lim_{h \to 0} \frac{f(x + h)g(x) - f(x)g(x + h)}{h g(x) g(x + h)} \\ &= \lim_{h \to 0} \frac{f(x + h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x + h)}{h g(x) g(x + h)} \\ &= \lim_{h \to 0} \frac{g(x)\frac{f(x + h) - f(x)}{h} - f(x)\frac{g(x + h) - g(x)}{h}}{g(x) g(x + h)} \\ &= \frac{g(x)f'(x) - f(x) g'(x)}{g(x)^2} \end{align}$$

This leads us to the so-called "quotient rule":

Which some people remember with the mnemonic "low D-high minus high D-low (over) square the low and away we go!"

Examples
The derivative of $$(4x - 2)/(x^2 + 1)$$ is:


 * $$\begin{align}

\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(x^2 + 1)(4) - (4x - 2)(2x)}{(x^2 + 1)^2} \\ &= \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} \\ &= \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2} \end{align}$$

Remember: the derivative of a product/quotient is not the product/quotient of the derivatives. (That is, differentiation does not distribute over multiplication or division.) However one can distribute before taking the derivative. That is $$\frac{d}{dx}\left((a+b)\times(c+d)\right) = \frac{d}{dx}\left(ac+ad+bc+bd\right) $$