Chain Rule Applied to Exponential Functions
Date: 10/11/2004 at 00:07:34 From: Brendan Subject: Chain Rule Word Problem At a time t hours after it was administered, the concentration of a drug in the body is f(t) = 27 e^(-.14t) ng/ml. What is the concentration 4 hours after it was administered? At what rate is the concentration changing at that time? I got lost finding the derivative of the problem to find the rate of change. Part 1. f(t) = 27 e^(-.14(4)) = 27e^(-.56) = 15.42 ng/ml Part 2. Chain rule = f'(g(x)) x g'(x) = 27'(e^-.56) x (e^-.56) I get lost after that. I don't know if I am going in the right direction and I think the derivative of 27 = 0, so the whole first half of the problem would equal 0.
Date: 10/11/2004 at 15:58:57 From: Doctor Mike Subject: Re: Chain Rule Word Problem Hi Brendan, The derivative of a constant times a function is just that constant, times the derivative of the function. So, the derivative of 27 e^(-.14t) is 27 times the derivative of e^(-.14t) . So, let's just concentrate on the derivative of e^(-.14t) and you can put it all together later. OK? People often have problems with the Chain Rule applied to exponentials, because of not being clear of what is the "outside" function and what is the "inside" function. That's why I like to use the notation exp(x) in place of the notation e^x when we do problems like this. Also, let's give a function name "h" to what is in your original exponent. That is, define it like h(t) = -.14t . Then, e^(-.14t) can be written as exp( h(t) ) which clearly shows that the exponential is the outside function, and what we have called "h" is the inside function. What do we do with this now? To use the Chain Rule you have to know how to differentiate both functions that are involved. The exponential function is its own derivative. exp'(t) = exp(t). You should have seen this already. For the other one, h'(t) = -.14 . That you should have seen a long time ago. Right? So, to use the Chain Rule on exp( h(t) ) you get exp'( h(t) ) * h'(t) which is exp( h(t) ) * (-0.14) . In this last expression, exp( h(t) ) is the derivative of the outside function evaluated at the inside function, and (-0.14) is the derivative of the inside function. The general notation for differentiating f(t) = g( h(t) ) with the C.R. is simply f'(t) = g'( h(t) ) * g'(t) . If you spend some time to get your function expressed in that way, then the rest will be easier. By the way, the reason why it is called the "chain" rule is that the chain can be more than just 2 links long. For instance, what if your main function can be defined using THREE other functions, like this. F(t) = G( H( K(t) ) ) You need to think of G as the outside function and H as the inside function, and apply the Chain Rule like this. F'(t) = G'( H( K(t) ) ) * H'( K(t) ) But what is H'( K(t) ) ??? That's the beauty of the chain rule, that you can keep adding on new "links" to the chain as long as you need to! You use the chain rule AGAIN on H to see that H'( K(t) ) = H( K(t) ) * K'(t) Here, as we are applying the chain rule a second time, H is the outside function at this stage and K is the inside function at this stage. EVERY time you use the chain rule you must be very clear what is the outside and inside function for that time. Finally let's combine both of those things to get the answer to my original problem, in the form of what you might call a "long chain." Namely, this: F'(t) = G'( H( K(t) ) ) * H'( K(t) ) = G'( H( K(t) ) ) * H( K(t) ) * K'(t) In harder problems you can easily have even longer chains. Hope this clarifies a few things and gets you back on track. I'm glad you wrote to Dr. Math. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/
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