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Mathematics LibreTexts

3.13: Multirule Derivatives

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Okay, let’s talk about ddx ex2+xsin(x). If you’re thinking this looks like a product rule, but it also looks like a chain rule, you’re right. To compute this derivative, we need to do the chain rule and the product rule. This is because it is a multirule problem. Let’s do this example

Multirule

Compute ddx ex2+xsin(x).

The way I like to break this down is to consider a little rule and a big rule. In this case, the little rule is the chain rule problem ddx ex2+x. If we do this problem, we see that f=ex, f=ex, g=x2+x and g=2x+1. So we have

ddx ex2+x=ex2+x(2x+1).

Now we are ready to do the big rule, which is the product rule. At this point we go back to the original problem ddx ex2+xsin(x). For this product rule, we see f=ex2+x, g=sin(x), g=cos(x). What is f? Why, that’s what we just computed in the equation above! So f=ex2+x(2x+1). Putting this all together with the product rule fg+gf, we have

ddx ex2+xsin(x)=fg+gf=ex2+xcos(x)+sin(x)ex2+x(2x+1).

Multirule

Compute ddxxsin(x2+x).

little chain rule: ddxsin(x2+x)

f=sin(x)g=x2+xf=cos(x)g=2x+1

Result: cos(x2+x)(2x+1)

Big quotient rule (aka the whole problem):ddxxsin(x2+x)

f=xg=sin(x2+x)f=1g=cos(x2+x)(2x+1)

Result: sin(x2+x)1xcos(x2+x)(2x+1)(sin(x2+x))2


This page titled 3.13: Multirule Derivatives 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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