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
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).
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)⋅1−xcos(x2+x)⋅(2x+1)(sin(x2+x))2