Quotient Rule

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We introduce the quotient rule as a way to decompose a quotient of functions when taking a derivative.

Video Lecture

Text and Additional Details

We’ve seen now that the sum and difference rules for derivatives are straight forward from limits, but the product rule didn’t follow exactly from limits. Nonetheless we were still able to deduce a reasonable (if not perfect) formula for the product rule. Next up is the quotient rule, which will be used when we want to take the derivative of a function of the form $\displaystyle \frac {f(x)}{g(x)}$ .

Developing the formula to deal with quotients will be a very similar process to the product rule. Let’s take a look at the difference quotient for $\frac {f(x)}{g(x)}$ to see why.

$\lim \limits _{\Delta x\rightarrow 0} \frac {\frac {f(x + \Delta x)}{g(x + \Delta x)} - \frac {f(x)}{g(x)}}{\Delta x} = \lim \limits _{\Delta x\rightarrow 0} \frac {\frac {f(x + \Delta x)g(x) - g(x + \Delta x)f(x)}{g(x + \Delta x)g(x)}}{\Delta x}$

Remember that in order to be differentiable at a point, the function must be continuous at that point; meaning that if $g(x) = 0$ then the function $\frac {f(x)}{g(x)}$ is not continuous and thus it is not differentiable. This means we may assume that $g(x)\neq 0$ for any point we need a quotient rule for... as otherwise trying to take derivative makes no sense. This means we can pull it out of the limit expression, which gives us the following:

$\displaystyle \lim \limits _{\Delta x\rightarrow 0} \frac {\frac {f(x + \Delta x)}{g(x + \Delta x)} - \frac {f(x)}{g(x)}}{\Delta x}$
$\displaystyle = \lim \limits _{\Delta x\rightarrow 0} \frac {\frac {f(x + \Delta x)g(x) - g(x + \Delta x)f(x)}{g(x + \Delta x)g(x)}}{\Delta x}$	Common Denominator
$\displaystyle = \left (\lim \limits _{\Delta x\rightarrow 0} \frac {1}{g(x + \Delta x)g(x)} \right ) \cdot \left (\lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x) - g(x + \Delta x)f(x)}{\Delta x}\right )$	Quotient Limit Law
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (\lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x) - g(x + \Delta x)f(x)}{\Delta x}\right )$	Evaluate first limit

Now, to continue the process, we will once again “add zero cleverly” by adding and subtracting $f(x)g(x)$ and factoring very much like we did while developing the product rule.

$\displaystyle \lim \limits _{\Delta x\rightarrow 0} \frac {\frac {f(x + \Delta x)}{g(x + \Delta x)} - \frac {f(x)}{g(x)}}{\Delta x}$
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (\lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x) - g(x + \Delta x)f(x)}{\Delta x}\right )$
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (\lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x)g(x) {\color {red} - f(x)g(x) + f(x)g(x)} - g(x + \Delta x)f(x)}{\Delta x}\right )$	Add and subtract $f(x)g(x)$
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (\lim \limits _{\Delta x\rightarrow 0} \frac {g(x) \bigg (f(x + \Delta x) - f(x)\bigg ) + f(x) \bigg ( g(x) - g(x + \Delta x)\bigg )}{\Delta x}\right )$	Regroup terms
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (\left (\lim \limits _{\Delta x\rightarrow 0} g(x) \frac {f(x + \Delta x) - f(x))}{\Delta x}\right ) + \left (\lim \limits _{\Delta x\rightarrow 0} f(x) \frac {g(x) - g(x + \Delta x)}{\Delta x}\right )\right )$	Split limit over addition sign
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (\left (g(x) \lim \limits _{\Delta x\rightarrow 0}\frac {f(x + \Delta x) - f(x))}{\Delta x}\right ) + \left (-f(x)\lim \limits _{\Delta x\rightarrow 0} \frac {g(x + \Delta x) - g(x)}{\Delta x}\right )\right )$	$f(x)$ and $g(x)$ are constant multipliers
	Since the limit uses $\Delta x$ not $x$ .
	Also factor out negative in second term.
$\displaystyle = \left ( \frac {1}{g(x)}\right )^2 \cdot \left (g(x) f'(x) - f(x)g'(x)\right )$	Replace Dif. Quot. with
	Prime Notation.

This gives us our quotient rule; in particular:

Quotient Rule Let $f(x)$ and $g(x)$ be differentiable functions. Then $\frac {d}{dx}\left [ \frac {f(x)}{g(x)} \right ] = \frac {g(x) f'(x) - f(x)g'(x)}{(g(x))^2}$ for any $x$ such that $g(x) \neq 0$ .

Notice here, unlike the product rule, that we have both division and subtraction, two operations that aren’t commutative... meaning that order matters now. This is typically remembered with the mnemonic “low D-high minus high D-low, all over low squared.”

However you remember it, it is key to remember which order the derivative and original functions appear. The first, or positive, term has the original denominator function $g(x)$ and the derivative of the numerator function, $f'(x)$ . The second term, or negative term, has the opposite; the original numerator function $f(x)$ and the derivative of the denominator function, $g'(x)$ .

So, again the algebra is pretty intense and it requires some clever manipulation, but the end result is a (relatively) compact rule for how to handle taking a derivative of quotients. It is worth a mention that, without these general rules, you would need to do similar algebra every time you had a quotient and had to do it with the difference quotient rather than one of these kinds of formulas. This should make us appreciate these derivative rules even more as they will save us countless hours of tedious computations in the future!

1 : One of the key differences of product and quotient derivative rules is...

Order of terms matters in the quotient rule, unlike the product rule. The product rule is simplier than the quotient rule. The product rule came first. There is no difference between the two rules, it’s all a conspiracy.