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)=gf′−fg′g2
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 ddx(b⋅ab). We set f=b, and g=ab. We see
ddx(b⋅ab)=fg′+g′f=b(ddxab)+(ab)(ddxb)=b(ddxab)+ab′b
But notice that ddx(b⋅ab)=ddxa=a′. Hence, we have
a′=b(ddxab)+ab′b
Now we just have to solve for ddxab, and we have a formula for derivatives of quotients!
a′=b(ddxab)+ab′ba′−ab′b=b(ddxab)1b(a′−ab′b)=ddxaba′b−ab′b2=ddxabba′b2−ab′b2=ddxabba′−ab′b2=ddxab
If we turn this equation around, it gives the same quotient rule I mentioned earlier:
ddx(ab)=ba′−ab′b2
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.
We set a=x and b=ex. We see a′=1, b′=ex. Using the formula ddx(ab)=ba′−ab′b2, we have
ddx(xex)=ba′−ab′b2=(ex)(1)−(x)(ex)(ex)2=ex(1−x)(ex)2=1−xex
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)=ba′−ab′b2=(x)(2x)−x2(1)(x)2=2x2−x2x2=x2x2=1
Same thing we got before!
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)=ba′−ab′b2=(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.
In this case a=sin(x) and b=x, so a′=cos(x) and b′=1. Hence
ddxsin(x)x=ba′−ab′b2=xcos(x)−sin(x)x2.