QuotienProduct Rule

The quotient rule can be used to take the derivative of two functions that are divided by each other. It is stated as

\[ (\frac{u}{v})’ = \frac{u’v - v’u}{v^2} \]

Proof

You can prove the quotient rule by substituting into the standard change in x formula that gives the derivatives for all differentiable functions.

\[ \frac{\frac{u(x + \Delta x)}{v(x + \Delta x)} - \frac{u(x)}{v(x)}}{\Delta x} \]

This is complicated expression, so the first step is to simplify the numerator by isolating it as

\[ \frac{u(x + \Delta x)}{v(x + \Delta x)} - \frac{u(x)}{v(x)} \]

We can then add in a \( u(x) \) and \( -u(x) \) to the first term of the expression resulting in a change of \( 0 \) but giving us more terms to work with, similar to the approach with the product rule proof.

\[ \frac{u(x) + u(x + \Delta x) - u(x)}{v(x) + v(x + \Delta x) - v(x)} - \frac{u(x)}{v(x)} \]

The expression is again becoming very complicated so we can simplify it again by defining some intermediary variables. These will be \( u \), \( v \), \( \Delta u \) and \( \Delta v \). They will mean

\[ u = u(x) \]

\[ v = v(x) \]

\[ \Delta u = u(x + \Delta x) - u(x) \]

\[ \Delta v = v(x + \Delta x) - v(x) \]

Substituting in again, we get a new expression, written in terms of variables. Don’t forget, you an always expand these variables to return to the expressions written in terms of \( x \).

\[ \frac{u + \Delta u}{v + \Delta v} - \frac{u}{v} \]

We can now create a common denominator for the expression by multiplying the denominators together and cross multiplying with the numerators.

\[ \frac{v(u + \Delta u) - u(v + \Delta v)}{v(v + \Delta v)} \]

Next multiply out the numerator, also called distribution.

\[ \frac{uv + v(\Delta u) - uv + u(\Delta v)}{v(v + \Delta v)} \]

The \( uv \) and \( -uv \) cancel.

\[ \frac{v(\Delta u) - u(\Delta v)}{v(v + \Delta v)} \]

Now that the numerator is simplified we can use it in our original expression.

\[ \frac{\frac{v(\Delta u) - u(\Delta v)}{v(v + \Delta v)}}{\Delta x} \]

Next, bring the divisor out as a multiplier.

\[ \frac{1}{\Delta x}\frac{v(\Delta u) - u(\Delta v)}{v(v + \Delta v)} \]

The multiplier can then we applied to a each of the numerator terms.

\[ \frac{v(\frac{\Delta u}{\Delta x}) - u(\frac{\Delta v}{\Delta x})}{v(v + \Delta v)} \]

We can now resolve are variables back to their original form.

\[ \frac{v(x)(\frac{u(x + \Delta x) - u(x)}{\Delta x}) - v(\frac{v(x + \Delta x) - v(x)}{\Delta x})}{v(x)(v(x) + v(\Delta x) - v(x))} \]

And deal with the divisor, in which the \( v(x) \) and \( -v(x) \) cancel to give

\[ \frac{v(x)(\frac{u(x + \Delta x) - u(x)}{\Delta x}) - v(\frac{v(x + \Delta x) - v(x)}{\Delta x})}{v(x)v(\Delta x)} \]

This now looks very similar to standard derivative definitions are the quotient rule definition. By writing with limits we know that.

\[ \lim_{x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} = u’(x) \]

\[ \lim_{x \to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} = v’(x) \]

\[ \lim_{x \to 0} u(x) = u(x) \]

\[ \lim_{x \to 0} v(x) = v(x) \]

\[ \lim_{x \to 0} v(x + \Delta x) = v(x) \]

Substituting this into the equation we get

\[ \lim_{x \to 0} \frac{v(x)v’(x) - v’(x)u(x)}{v(x)v(x)} \]

Which is equivalent to

\[ (\frac{u}{v})’ = \frac{u’v - v’u}{v^2} \]

Thus proving the quotient rule.