Calculus Chain RuleDate: 11/23/97 at 20:09:15 From: stuart Subject: Calculus chain rule I can't understand the chain rule. Every time I ask someone to explain it they use y's and u's, etc... could you give me the chain rule in easy terms, like how to do it, not just give me a formula like y=(U)^2? Thanks. Stu Date: 11/23/97 at 21:24:09 From: Doctor Scott Subject: Re: Calculus chain rule Hi Stu! Good question. I was just skimming an excellent Calculus book written by Paul Foerster where this very question was addressed. His suggestion was that you should think of the chain rule as a process rather than a rule with a lot of du/dx and dy/dx's. So, here goes.... Remember that the chain rule is used to find the derivative of *compositions of functions* - that is, functions that have functions inside of them. For example, the function sin(x^2) can be thought of as a composition of two other functions, sin x and x^2, with the x^2 being INSIDE the sin function. Similarly, the function (x^2 - 5x + 8)^(1/2) is also a composition of two other functions, (x^2 - 5x + 8) and x^(1/2), with the first function being INSIDE the second. One more example? The function cos(tan(5x-3)) is the composition of three functions, 5x - 3 inside of tan x, inside of cos x. So the chain rule gets applied when there is some function INSIDE of another function. The stuff that people have been telling you probably goes something like this: If y = sin(4x-3), then we can write this function as the composition of y = sin u and u = 4x - 3. (Again, notice that the 4x - 3 is INSIDE of the sin function.) Then, dy/dx = dy/du * du/dx. So, we have dy/dx = cos u * 4; but u = 4x-3, so we have dy/dx = 4cos(4x-3). How about another way? Let's think of the chain rule as a process. The derivative of a composite function is the DERIVATIVE OF THE OUTSIDE FUNCTION TIMES the DERIVATIVE OF THE INSIDE FUNCTION. In practice, here's how it works. Consider y = sin(4x-3). The outside function is a sine function; its derivative is cosine, so we have (so far) cos(4x-3). Now, INSIDE the sine function is 4x-3. Its derivative is 4, so now we have 4cos(4x-3). Notice that there is no other function "inside" the 4x-3, so we are done. Let's look at a couple more examples: y = (x^2 - 5x + 8)^(1/2). The OUTSIDE FUNCTION is basically a power rule problem, so we have 0.5(x^2 - 5x + 8)^(-1/2) using the power rule. The INSIDE FUNCTION is x^2 - 5x + 8; its derivative is 2x - 5, so we have y' = (2x - 5)(.5)(x^2 - 5x + 8)^(-1/2). y = cos(tan(5x-3)). The outermost function is a cosine, so its derivative is negative sine: -sin(tan(5x-3)). Inside the cosine is a tan function; its derivative is sec^2, so we now have sec^2 (5x-3) * (-sin(tan(5x-3)) Finally, inside of the tan function is 5x-3; its derivative is 5. So, FINALLY, we have 5 * sec^2 (5x-3) * (-sin(tan(5x-3)) Or, simplifying, we get y' = -5 sec^2 (5x-3) sin(tan(5x-3)) So, it helps a lot to think of the chain rule as: The derivative of the outside TIMES the derivative of what's inside! -Doctor Scott, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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