Proof of the Quotient Rule

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

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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.

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

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