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Calculus Chain Rule


Date: 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/   
    
Associated Topics:
High School Calculus

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