3.9: Quotient Rule

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

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