# 3.9: Quotient Rule

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What about $$\frac{d}{dx} \left( \frac{x}{e^x} \right)$$? Can we just take the derivative of the top and bottom separately, and put them together? Nope, we need a quotient rule.

$$\boxed{\cfrac{d}{dx} \left( \cfrac{f}{g} \right) = \frac{g f' - f g'}{g^2}}$$

Where did this strange formula come from? Some fancy algebra will get you there as the next example shows.

## Proving the Quotient Rule

Prove the quotient rule $$\cfrac{d}{dx} \left( \cfrac{a}{b} \right) = \frac{b a' - ab'}{b^2}$$.

Note that I’m using $$a$$ and $$b$$ instead of $$f$$ and $$g$$ right now because we will need $$f$$ and $$g$$ to mean something else in just a second.

So what can we do? One way it to use the product rule in a strange manner. We are going to apply it to $$\frac{d}{dx} \left( b \cdot \frac{a}{b}\right)$$. We set $$f = b$$, and $$g = \frac{a}{b}$$. We see

\begin{align*} \frac{d}{dx} \left( b \cdot \frac{a}{b} \right ) & = f g' + g' f \\ & = b \left( \frac{d}{dx} \frac{a}{b} \right) + \left(\frac{a}{b} \right) \left( \frac{d}{dx} b \right) \\ & = b \left( \frac{d}{dx} \frac{a}{b} \right) + \frac{a b'}{b} \end{align*}

But notice that $$\frac{d}{dx} \left(b \cdot \frac{a}{b}\right) = \frac{d}{dx} a = a'$$. Hence, we have

$$a' = b \left( \frac{d}{dx} \frac{a}{b}\right) + \frac{a b'}{b}$$

Now we just have to solve for $$\frac{d}{dx} \frac{a}{b}$$, and we have a formula for derivatives of quotients!

\begin{align*} a' & = b \left( \frac{d}{dx} \frac{a}{b} \right) + \frac{a b'}{b} \\ a' - \frac{a b'}{b} & = b\left( \frac{d}{dx} \frac{a}{b} \right) \\ \frac{1}{b} \left( a' - \frac{a b'}{b} \right) & = \frac{d}{dx} \frac{a}{b} \\ \frac{a'}{b} - \frac{a b'}{b^2} & = \frac{d}{dx} \frac{a}{b} \\ \frac{b a'}{b^2} - \frac{a b'}{b^2} & = \frac{d}{dx} \frac{a}{b} \\ \frac{b a' - a b'}{b^2} & = \frac{d}{dx} \frac{a}{b} \\ \end{align*}

If we turn this equation around, it gives the same quotient rule I mentioned earlier:

$$\boxed{\cfrac{d}{dx} \left( \cfrac{a}{b} \right) = \frac{b a' - ab'}{b^2}}$$

This has a cute rhyme to it: “low dee high minus high dee low, over the square of what’s below”. The “low dee high” means $$b a'$$, since $$b$$ is the “low” and $$a'$$ is the “dee high”. Then “minus high dee low” is $$- a b'$$. Finally, “over the square of what’s below” is $$b^2$$.

Let’s see how it looks applying the quotient rule.

## Quotient Rule with $$\frac{x}{e^x}$$

Find $$\frac{d}{dx} \left ( \frac{x}{e^x} \right)$$.

We set $$a = x$$ and $$b = e^x$$. We see $$a' = 1$$, $$b' = e^x$$. Using the formula $$\cfrac{d}{dx} \left( \cfrac{a}{b} \right) = \frac{b a' - ab'}{b^2}$$, we have

\begin{align*} \frac{d}{dx} \left ( \frac{x}{e^x} \right) & = \frac{b a' - ab'}{b^2} \\ & = \frac{(e^x)(1) - (x)(e^x)}{(e^x)^2} \\ & = \frac{e^x (1 - x)}{(e^x)^2} \\ & = \boxed{\frac{1 - x}{e^{x}}} \end{align*}

## Quotient Rule with $$\frac{x^2}{x}$$

Find $$\frac{d}{dx} \left( \frac{x^2}{x} \right)$$ using the quotient rule and power rule.

If we simplify and turn $$\frac{x^2}{x}$$ into just $$x$$, then we have $$\frac{d}{dx} \left( \frac{x^2}{x} \right) = \frac{d}{dx} x = \boxed{1}$$. Easy enough.

Using the quotient rule, we set $$a = x^2$$ and $$b = x$$, with $$a' = 2x$$ and $$b = 1$$. We have

\begin{align*} \frac{d}{dx} \left( \frac{x^2}{x} \right) & = \frac{b a' - a b'}{b^2} \\ & = \frac{(x)(2x) - x^2(1)}{(x)^2} \\ & = \frac{2x^2 - x^2}{x^2} \\ & = \frac{x^2}{x^2} \\ & = \boxed{1} \end{align*}

Same thing we got before!

## More Quotient Rule

Find $$\frac{d}{dx} \frac{(x^2 + 2x)}{\ln(x)}$$.

In this case, $$a = x^2 + 2x$$ and $$b = \ln(x)$$. We have $$a' = 2x + 2$$, $$b' = \frac{1}{x}$$. Hence we have

\begin{align*} \frac{d}{dx} \frac{(x^2 + 2x)}{\ln(x)} & = \frac{b a' - a b'}{b^2} \\ & = \frac{(\ln(x)) (2x + 2) - (x^2 + 2x) \left( \frac{1}{x} \right)}{(\ln x)^2} \\ & = \boxed{\frac{\ln(x)(2x + 2) - (x + 2)}{(\ln x)^2}} \end{align*}

This doesn’t really simplify farther, so that’s our answer.

## With a sine this time

Find $$\frac{d}{dx} \frac{\sin(x)}{x}$$.

In this case $$a = \sin(x)$$ and $$b = x$$, so $$a' = \cos(x)$$ and $$b' = 1$$. Hence

\begin{align*} \frac{d}{dx} \frac{\sin(x)}{x} & = \frac{b a' - a b'}{b^2} \\ & = \boxed{\frac{x \cos(x) - \sin(x)}{x^2}}. \end{align*}

This page titled 3.9: Quotient Rule is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.