 ## Proof of the Quotient Rule to accompany Calculus Applied to the Real World

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Español The Quotient Rule

If the functions $f$ and $g$ are differentiable at $x,$ with $g(x) ≠ 0,$ then the quotient $\frac{f}{g}$ is differentiable at $x,$ and

 $\frac{d}{dx}\ \Bigl$\frac{f}{g} \Bigr$\ (x) = \frac{f^{'} (x) g(x) - f(x) g^{'}(x)}{[g(x)]^2.}$

Proof By the definition of the derivative,

 $\frac{d}{dx} \left$\frac{f}{g} \right$ (x)$ $=$ $\lim_{h \to 0} \frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)} h$ $=$ $\lim_{h \to 0} \frac{f(x+h)g(x) - f(x)g(x+h)}{g(x+h)g(x)h}$ (subtraction of fractions) $=$ $\lim_{h \to 0} \frac{f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h)}{g(x+h)g(x)h}$ (we added & subtracted $f(x)g(x)$) $=$ $\lim_{h \to 0} \frac{[f(x+h) - f(x)]g(x) - f(x)[(g(x+h) - g(x)]}{g(x+h)g(x)h}$ (a little algebra) $=$ $\lim_{h \to 0} \frac{\frac{f(x+h) - f(x)}{h} g(x) - f(x) \frac{g(x+h) - g(x)}{h}}{ g(x+h)g(x)}$ (a little more algebra)

If we recognize the difference quotients for $f$ and $g$ in this last expression, we see that taking the limit as $h \to 0$ replaces them by the derivatives $f^{'}(x)$ and $g^{'}(x).$ Further, since $g$ is differentiable, it is also continuous, and so $g(x+h) \to g(x)$ as $h \to 0.$ Putting this all together gives
$\frac{d}{dx}\ \Bigl$\frac{f}{g}\ \Bigr$ (x) = \frac{f^{'}(x) g(x) - f(x) g^{'}(x)}{[g(x)]^2},$

which is the quotient rule. 