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3.6 chain rule.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Tues 9/30/08 3.6 - The Chain Rule And now, the chain rule! The chain rule comes into play whenever we want to take the derivative d of a composition of functions. In other words, how would we go about calculating [f(g(x))]? dx First, let’s replace g(x) with u. It’s a little nicer to see this way. So, we want to eventually d 0 calculate dx [f(u)]. Well, this isn’t just f (u), because u is also a function of x, and we can’t just d go ahead and do something like du[f(u)], because we want the derivative with respect to x. du Hmm...but what if we combine the two? So, we also have another derivative thrown in there, . dx Now think of the \du" as a variable (I know it’s not, but just pretend for a second), and if we multiply the two together, we can cancel out the du’s. In other words, dy dy du = ▯ dx du dx Bringing this all back to compositions of functions, we get: Chain Rule: d [f(g(x))] = f (g(x))g (x) dx Or, in other words, it’s the derivative of the outside function with the inside function plugged in times the derivative of the inside function. Example: f(x) = (x + 1) . Find f (x).0 Solution: f (x) = 5(x + 1) ▯ (2x) 2 4 The 5(x + 1) comes from the derivative of the outside function with the inside one plugged in, and then the 2x is the derivative of what’s inside. Example: f(x) = sin(x + 2x ▯ 5)2 Solution: 0 2 3 2 f (x) = (3x + 4x)cos(x + 2x ▯ 5) Guess what! You can derive the Quotient Rule a di▯erent way by using the product rule and chain rule! Well, now that we’ve been introduced to the chain rule, let’s check it out! d f(x) d ▯1 = f(x)[g(x)] dx g(x) dx = f (x)[g(x)] ▯1 + f(x)(▯1)[g(x)] ▯2g (x) f (x)g(x) f(x)g(x) = 2 ▯ 2 g (x) g (x) gf ▯ fg 0 = g2 1 A much nicer way to ▯gure out the quotient rule rather than using the de▯nition of the derivative, and by all means, if you can’t remember the quotient rule, you can always come back to something like this! Let’s do a few more examples. Example: f(x) = (sin(x)cos(x)) . This problem would suck if you had to do product rule (three times!!!). Solution: f (x) = 2(sin(x)cos(x))[cos(x)cos(x) + sin(x)(▯sin(x))] = 2sin(x)cos (x) ▯ 2cos(x)sin (x) 3 Much nicer than doing the product rule 3 times! What do you think might happen if we have the composition of the composition of functions? Ie, what’s d f(g(h(x)))? dx Just work with it one function at a time. Chain rule says derivative of the outside with the inside plugged in, times the derivative of the inside. So, d f(g(h(x))) = f (g(h(x)))g (h(x))h (x) 0 dx Example: f(x) = sin(cos(x )) 2 Solution:
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