Calculus/Quotient Rule

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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":

Derivatives of quotients (Quotient Rule)

 \frac{d}{dx} \left[{f(x)\over g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(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[edit]

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)