Quotient rule

The quotient rule is used to determine the derivative of a function expressed as the quotient of 2 differentiable functions. It is defined as shown:

${d\over dx}{u\over v}={v{du\over dx}-u{dv\over dx}\over v^2}$

Also written as:

$f'(x)={vu'-uv'\over v^2}$

This can also be done as a Product rule (with an inlaid Chain rule):

${d\over dx} (u v^{-1}) = u(-1v^{-2} v') + (v^{-1})u'$ ${d\over dx} (u v^{-1}) = {-uv'\over v^{2}} + {u'\over v}$ ${d\over dx} (u v^{-1}) = {u'v-uv'\over v^{2}}$

You may do this whichever way you prefer.

Proof

We know that the two following limits exist as $f(x), g(x)$ are differentiable. We also have the condition that $g(x) \ne 0$.

$f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x+h) -f(x)}{h}$

$g'(x) = \lim_{h \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

Applying the first principles definition of differentiation we get

$\left(\dfrac{f(x)}{g(x)}\right)' = \lim_{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{f(x+h)}{g(x+h)} - \dfrac{f(x)}{g(x)}\right)$

We can combine the two fractions into one fraction by cross-multiplying.

$\left(\dfrac{f(x)}{g(x)}\right)' = \lim_{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{f(x+h)g(x) - g(x+h)f(x)}{g(x+h)g(x)}\right)$

If we add and subtract a $f(x) g(x)$, it will change nothing.

$\left(\dfrac{f(x)}{g(x)}\right)' = \lim_{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{f(x+h)g(x) -f(x)g(x) + f(x)g(x) - g(x+h)f(x)}{g(x+h)g(x)}\right)$

If we bring the $g(x+h)g(x)$ term in the denominator to the front, and separate the numerator with algebra of limits we will have

$\left(\dfrac{f(x)}{g(x)}\right)' = \lim_{h \rightarrow 0} \dfrac{1}{g(x+h)g(x)} \left( \dfrac{f(x+h)g(x) -f(x)g(x)}{h} - \dfrac{g(x+h)f(x) - f(x)g(x)}{h}\right)$

We now break this into a product of two limits, and a sum of two limits.

$\left(\dfrac{f(x)}{g(x)}\right)' = \lim_{h \rightarrow 0} \dfrac{1}{g(x+h)g(x)} \cdot \left(g(x) \lim_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h} - f(x)\lim_{h \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}\right)$

Evaluating all limits yields

$\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{1}{g(x)^2} \cdot (f'(x)g(x) - g'(x)f(x))$

Putting this in a more familiar form, we have

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

This completes the proof.