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:

$\begin{array}{cc}\frac{d}{dx}\frac{f(x)}{g(x)}& =\underset{h\to 0}{\lim }\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}\\ & =\underset{h\to 0}{\lim }\frac{f(x+h)g(x)-f(x)g(x+h)}{hg(x)g(x+h)}\\ & =\underset{h\to 0}{\lim }\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{hg(x)g(x+h)}\\ & =\underset{h\to 0}{\lim }\frac{g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h}}{g(x)g(x+h)}\\ & =\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{g(x{)}^2}\end{array}$

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

Derivatives of quotients (Quotient Rule) $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{g(x{)}^2}$

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:

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

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 $\frac{d}{dx}((a+b)\times (c+d))=\frac{d}{dx}(ac+ad+bc+bd)$