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.
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.
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
\[\begin{align*} \frac{d}{dx} \left ( \frac{x}{e^x} \right) & = \frac{b a' - ab'}{b^2} \\ & = \frac{(e^x)(1) - (x)(e^x)}{(e^x)^2} \\ & = \frac{e^x (1 - x)}{(e^x)^2} \\ & = \boxed{\frac{1 - x}{e^{x}}} \end{align*}\]
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!
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.
In this case \(a = \sin(x)\) and \(b = x\), so \(a' = \cos(x)\) and \(b' = 1\). Hence
\[\begin{align*} \frac{d}{dx} \frac{\sin(x)}{x} & = \frac{b a' - a b'}{b^2} \\ & = \boxed{\frac{x \cos(x) - \sin(x)}{x^2}}. \end{align*}\]