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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 

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 
Associated Topics:
College Calculus
High School Calculus

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