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3.9: Quotient Rule

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What about ddx(xex)? Can we just take the derivative of the top and bottom separately, and put them together? Nope, we need a quotient rule.

ddx(fg)=gffgg2

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 ddx(ab)=baabb2.

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 ddx(bab). We set f=b, and g=ab. We see

ddx(bab)=fg+gf=b(ddxab)+(ab)(ddxb)=b(ddxab)+abb

But notice that ddx(bab)=ddxa=a. Hence, we have

a=b(ddxab)+abb

Now we just have to solve for ddxab, and we have a formula for derivatives of quotients!

a=b(ddxab)+abbaabb=b(ddxab)1b(aabb)=ddxabababb2=ddxabbab2abb2=ddxabbaabb2=ddxab

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

ddx(ab)=baabb2

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 ba, since b is the “low” and a is the “dee high”. Then “minus high dee low” is ab. Finally, “over the square of what’s below” is b2.

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

Quotient Rule with xex

Find ddx(xex).

We set a=x and b=ex. We see a=1, b=ex. Using the formula ddx(ab)=baabb2, we have

ddx(xex)=baabb2=(ex)(1)(x)(ex)(ex)2=ex(1x)(ex)2=1xex

Quotient Rule with x2x

Find ddx(x2x) using the quotient rule and power rule.

If we simplify and turn x2x into just x, then we have ddx(x2x)=ddxx=1. Easy enough.

Using the quotient rule, we set a=x2 and b=x, with a=2x and b=1. We have

ddx(x2x)=baabb2=(x)(2x)x2(1)(x)2=2x2x2x2=x2x2=1

Same thing we got before!

More Quotient Rule

Find ddx(x2+2x)ln(x).

In this case, a=x2+2x and b=ln(x). We have a=2x+2, b=1x. Hence we have

ddx(x2+2x)ln(x)=baabb2=(ln(x))(2x+2)(x2+2x)(1x)(lnx)2=ln(x)(2x+2)(x+2)(lnx)2

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

With a sine this time

Find ddxsin(x)x.

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

ddxsin(x)x=baabb2=xcos(x)sin(x)x2.


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.

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