# How do you prove the quotient rule?

Oct 2, 2014

By the definition of the derivative,

$\left[\frac{f \left(x\right)}{g \left(x\right)}\right] ' = {\lim}_{h \to 0} \frac{f \frac{x + h}{g} \left(x + h\right) - f \frac{x}{g} \left(x\right)}{h}$

by taking the common denominator,

$= {\lim}_{h \to 0} \frac{\frac{f \left(x + h\right) g \left(x\right) - f \left(x\right) g \left(x + h\right)}{g \left(x + h\right) g \left(x\right)}}{h}$

by switching the order of divisions,

$= {\lim}_{h \to 0} \frac{\frac{f \left(x + h\right) g \left(x\right) - f \left(x\right) g \left(x + h\right)}{h}}{g \left(x + h\right) g \left(x\right)}$

by subtracting and adding $f \left(x\right) g \left(x\right)$ in the numerator,

$= {\lim}_{h \to 0} \frac{\frac{f \left(x + h\right) g \left(x\right) - f \left(x\right) g \left(x\right) - f \left(x\right) g \left(x + h\right) + f \left(x\right) g \left(x\right)}{h}}{g \left(x + h\right) g \left(x\right)}$

by factoring $g \left(x\right)$ out of the first two terms and $- f \left(x\right)$ out of the last two terms,

$= {\lim}_{h \to 0} \frac{\frac{f \left(x + h\right) - f \left(x\right)}{h} g \left(x\right) - f \left(x\right) \frac{g \left(x + h\right) - g \left(x\right)}{h}}{g \left(x + h\right) g \left(x\right)}$

by the definitions of $f ' \left(x\right)$ and $g ' \left(x\right)$,

$= \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{\left[g \left(x\right)\right]}^{2}}$

I hope that this was helpful.