2.5 The Chain Rule Brian E. Veitch
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1 2.5 The Chain Rule This is our last ifferentiation rule for this course. It s also one of the most use. The best way to memorize this (along with the other rules) is just by practicing until you can o it without thinking about it. Ok, so what s the chain rule? It s the rule that allows us to ifferentiate a composition of functions. One way to think about a composition of functions to think of it having an outsie function an insie function. For example, 1. y = (3x 8 + x) 25 (a) The outsie function: f(x) = x 25 (b) The insie function: g(x) = 3x 8 + x To check this, what is f g? f g = f (g(x)) = f ( 3x 8 + x ) = ( 3x 8 + x ) y = x (a) The outsie function: f(x) = x 127
2 (b) The insie function: g(x) = x (c) f (g(x)) = x y = cos 4 (x) (a) Remember that cos 4 (x) is a special notation. It represents (cos(x)) 4 (b) The outsie function: f(x) = x 4 (c) The insie function: g(x) = cos(x) () f (g(x)) = (cos(x)) 4 4. y = sec 2 ( (x)) (a) A composition of functions oesn t have to be just a composition of two functions. In this example, it s a composition of three functions. (b) Rewrite y as y = (sec( x)) 2 (c) The outsie function: f(x) = x 2 () 1st insie function: g(x) = sec(x) (e) 2n insie function: h(x) = x 128
3 (f) The composition is y = f (g (h(x))) We went through all those examples because it s important you know how to ientify the composition. What is the outsie function? What is the insie function? Is there another insie function? Chain Rule If g is ifferentiable at x an f is ifferentiable at g(x), then the composite function f (g(x)) is ifferentiable an [f (g(x))] = f (g(x)) g (x) Steps: 1. Differentiate the outsie function f 2. Leave the insie function alone g 3. Multiply by the erivative of the insie function g Example Fin y when y = (x 3 + 5x) 5 1. Ientify the outsie function f an the insie function g. 129
4 (a) Outsie function: f(x) = x 5, with f (x) = 5x 4 (b) Insie function: g(x) = x 3 + 5x, with g (x) = 3x Write out the chain rule formula. y = f (g(x)) g (x) 3. Fill in the parts (You won t be oing it like this once you get better at it) y = f (g(x)) g (x) = 5 (g(x)) 4 g (x) = 5 ( x 3 + 5x ) 4 ( 3x ) Example If y = cos 2 (x), fin y. 1. Rewrite y as y = (cos(x)) 2 2. Ientify the outsie an insie functions. (a) The outsie function: f(x) = x 2, with f (x) = 2x (b) The insie function: g(x) = cos(x), with g (x) = sin(x) 130
5 3. Use the chain rule y = 2 (cos(x)) sin(x) y = 2 cos(x) sin(x) Example If y = sec(5x), fin y. You may be surprise that many stuents get this wrong. It s because of the erivative of sec(x). Let s get starte. 1. Ientify the outsie an insie functions (a) The outsie function: f(x) = sec(x), with f (x) = sec(x) tan(x). (b) The insie function: g(x) = 5x, with g (x) = Use the Chain Rule y = f (g(x)) g (x) = sec (g(x)) tan (g(x)) g (x) = sec(5x) tan(5x) 5 = 5 sec(5x) tan(5x) Can you fin y? It s not that easy. Notice that it is now a prouct of functions. This means you ll have to o the prouct rule an the chain rule in the same problem. 131
6 Example Fin [cos(x 5 + sin(x))] 1. Ientify the outsie an insie functions (a) The outsie function: f(x) = cos(x), with f (x) = sin(x) (b) The insie function: g(x) = x 5 + sin(x), with g (x) = 5x 6 + cos(x) 2. When I o the chain rule, I say the following in the hea, (a) Differentiate the outsie function an leave the insie alone (b) Multiply by the erivative of the insie 3. Use the chain rule y = sin ( x 5 + sin(x) ) ( 5x 6 + cos(x) ) So far we ve ifferentiate a composition of two functions. Recall that a composition of functions can have any number of functions. Here s how the chain rule looks when you have a composition of three functions. 132
7 2.5.2 Chain Rule with a composition of three functions [f (g (h(x)))] = f (g (h(x))) g (h(x)) h(x) Example Fin y when y = cos 5 ( x). 1. Anytime you re aske to ifferentiate a trig function with an exponent like this, rewrite it. y = ( cos( x) ) 5 2. Ientify the insie an outsie functions. I on t know if you can tell yet, but there s more than one insie function. (a) The outsie function: f(x) = x 5, with f (x) = 5x 4 (b) The insie function: g(x) = cos( x). Note that g (x) requires the chain rule. 3. Use the chain rule y = 5(cos( x) 4 [ ] cos( x) = 5(cos( x) 4 sin( [ ] x) x = 5(cos( x) 4 sin( x) 1 2 x 1/2 = 5 2 x 1/2 (cos( x)) 4 sin( x) 133
8 Now it s time for some more complicate erivatives. prouct, an quotient rule at the same time. These involve using the chain, Example Fin y when y = (cos(x) + x3 ) 4. tan(x 2 ) 1. Notice that when we ifferentiate the numerator (cos(x) + x 3 ) an the enominator tan(x 2 ), we will use the chain rule. But the overall function is a quotient. So which one goes first? 2. Since the main function is a quotient, we use the quotient rule. 3. Let s get starte with the quotient rule. y = tan(x 2 ) [(cos(x) + x3 ) 4 ] (cos(x) + x 3 ) 4 [tan(x2 )] (tan(x 2 )) 2 4. I STRONGLY recommen showing all your work. Notice that all I have to o now is fill in those with the correct erivatives. Breaking it up like this is very useful. [ (cos(x) + x 3 ) 4] = 4 ( cos(x) + x 3) 3 [ ] cos(x) + x 3 = 4 ( cos(x) + x 3) 3 ( ) sin(x) + 3x 2 [ tan(x 2 ) ] = sec 2 (x 2 ) = sec 2 (x 2 ) 2x [ x 2 ] 134
9 5. Now just replace the s an we re one. y = = tan(x 2 ) [(cos(x) + x3 ) 4 ] (cos(x) + x 3 ) 4 [tan(x2 )] (tan(x 2 )) 2 tan(x 2 ) [ ] 4 (cos(x) + x 3 ) 3 ( sin(x) + 3x 2 ) (cos(x) + x 3 ) 4 [sec 2 (x 2 ) 2x] (tan(x 2 )) 2 Example Fin y when y = (x 3 2x 11 ) 4 (1 15x 2 ) The overall function is a prouct of two functions (x 3 2x 11 ) 4 an (1 15x 2 ) 100. We ll start with the prouct rule. 2. Prouct Rule y = (x 3 2x 11 ) 4 [ (1 15x 2 ) 100] + (1 15x 2 ) 100 [ (x 3 2x 11 ) 4] 3. We have two chain rules to work on now. (a) [(1 15x2 ) 100 ] [ (1 15x 2 ) 100] = 100(1 15x 2 ) 99 [ ] 1 15x 2 = 100(1 15x 2 ) 99 ( 30x) 135
10 (b) [(x3 2x 11 ) 4 ] [ (x 3 2x 11 ) 4] = 4(x 3 2x 11 ) 3 [ x 3 2x 11] = 4(x 3 2x 11 ) 3 (3x 2 22x 10 ) 4. Now put it all together y = (x 3 2x 11 ) 4 [ (1 15x 2 ) 100] + (1 15x 2 ) 100 [ (x 3 2x 11 ) 4] = (x 3 2x 11 ) 4 [100(1 15x 2 ) 99 ( 30x) ] + (1 15x 2 ) 100 [4(x 3 2x 11 ) 3 (3x 2 22x 10 ) ] You have some common factors you can pull out. I ll leave that to you as an exercise. At some point in the near future, you will have to know how to factor this. Example Fin y x when y = x 3 2 cos(x) 1. Make sure everything is written in the correct form. ( x y = x 3 2 cos(x) ) 1/2 2. This is a chain rule first, then a quotient rule. 3. Start the chain rule 136
11 y = 1 ( ) x 2 1/ x 3 2 cos(x) 4. So now we just have to fin [ ] x x 3 2 cos(x) [ x x 3 2 cos(x) ] [ x ] x 3 2 cos(x) = (x3 2 cos(x)) (2x) (x 2 + 1) (3x sin(x)) (x 3 2 cos(x)) 2 5. Put it all together y = 1 ( ) x 2 1/ x 3 2 cos(x) = 1 ( ) x 2 1/ x 3 2 cos(x) [ ] x x 3 2 cos(x) [ ] (x 3 2 cos(x)) (2x) (x 2 + 1) (3x sin(x)) (x 3 2 cos(x)) 2 Again, I m not focuse with simplifying at the moment. You will have the rest of the semester simplifying these things. Plus, I m also giving examples that are nasty to simplify. Example If h(x) = g(f(x)), m(x) = g(x 3 ), f(2) = 8, g (8) = 2, an f (2) = 3, fin h (2). 1. Fin h (x) using the chain rule h (x) = g (f(x)) f (x) 137
12 2. Plug in x = 2 h (2) = g (f(2)) f (2) = g (8) 3 = 2 3 = 6 Example Using the previous example s information, fin m (2). 1. Fin m (x) m (x) = g (x 3 ) [ ] x 3 = g (x 3 ) 3x 2 2. Plug in x = 2 m (2) = g (2 3 ) 3(2) 2 = g (8) 12 = 2 12 = 24 As I state before, practice these rules. In your textbook there are plenty of problems to practice. 138
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