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Trigonometry and the Chain Rule


Date: 12/01/98 at 17:59:54
From: Amanda
Subject: Calculus - the trig unit

I have three questions that have me stumped. I need to differentiate 
the following:

   y = 2 csc^3(sqrt(x))
   y = x/2 - (sin (2x))/4
   y = (1 - cos (x))/sin (x)


Date: 12/01/98 at 19:37:16
From: Doctor Santu
Subject: Re: Calculus - the trig unit

Amanda:

These all have to do with the Chain Rule. Here's the basic idea. 
Suppose:

   y = sin (x^3 + tan (x)).

How do you find the derivative?

Think of x^3 + tan x as a big BLOB. So we really need to find the 
derivative of:

   y = sin (BLOB)

Well, the rule says that the derivative of sin (BLOB) is simply 
cos (BLOB) multiplied by the derivative of the BLOB itself.

A word on notation: I'm going to write y' for the derivative of y 
(instead of dy/dx).

Now, in this case:

   y' = cos (x^3 + tan(x)) * (3x^2  + sec^2 (x))

because BLOB is x^3 + tan (x), and the derivative of the BLOB is 
3x^2 + sec^2 (x).  

Let's try another example:

   y = sin^3 (x^3 + tan x)

This is really:

   y = [ sin (x^3  + tan x) ]^3

Using our previous terminology, the derivative of (blob)^3 is simply:

   3(blob)^2 * (the derivative of blob itself).

In this case:

   y' = 3[sin (x^3 + tan (x))]^2 * (derivative of sin (x^3 + tan(x)))
      = 3[sin(x^3 + tan (x))]^2 * cos (x^3 + tan (x)) * 
        (derivative of x^3 + tan (x))
      = 3[sin(x^3 + tan(x))]^2 * cos(x^3 + tan(x)) * (3x^2 + sec^2(x))

For another example:

   y = csc^3 (x^3 + tan (x))

It's the same as:

   y = [csc (x^3 + tan (x))]^3

So:

   y' = 3[csc(x^3 + tan (x))]^2 * (deriv. of csc (x^3 + tan (x)) )
      = 3[csc(x^3 + tan (x))]^2 * (-csc(x^3 + tan (x)) * 
        cot(x^3 + tan (x)) * (x^3 + tan (x))')
      = 3[csc(x^3 + tan (x))]^2 * (-csc(x^3 + tan (x)) * 
        cot(x^3 + tan (x)) * (3x^2+sec^2 (x)))

For example 3, try:

   y = csc^3( sqrt (x) )

Remember it's simply:

   y = [csc(sqrt (x))]^3

So it should start off with:

   3[csc(sqrt (x))]^2 * (the derivative of the cosecant...)

To take the derivative of [sin (2x)]/4, just rewrite it as:

   (1/4) * sin (2x)  

The 1/4 is just a constant, so it just sits there, and the derivative 
would be:

   (1/4) * cos (2x) * (2) = (1/2) * cos (2x)

In the chain rule, the basic idea is to peel the onion from the 
outside. You want to take the derivative of a function within a 
function within a function. You take the derivative of the outermost 
function relative to the stuff that's inside it, then multiply that by 
the derivative of the inside expression, relative to the expression 
inside the expression, and so on, all the way down to the tiniest 
little x all the way inside. (And some people even stick a "1" on at 
the end, because the derivative of an x is just 1. I think that's 
overdoing it a bit.)

The Chain Rule needs quite a lot of imagination to see these formulas 
as expressions within expressions, and ideally you should have a friend 
sit by you and point out how to "peel the onion" layer by layer.

One final example:

   y = sin (tan (sin^2 (x^7 + 3x) ) )

   y' = ...?  

You must first take the derivative of sin (expression), relative to the 
expression that's inside. You multiply that by the derivative of the 
tan (inside expression). You multiply that by the derivative of 
[sin (x^7 + 3x)]^2, because sin^2 (x^7+3x) means [sin(x^7+3x)]^2. 
That, in turn, will contain the derivative of sin(x^7 + 3x), which in 
turn will contain the derivative of (x^7 + 3x), which is 7x^6 + 3.

It's important to put the proper expression inside the various partial 
expressions. So:

   y' = cos(tan ...) * sec^2(sin ...) * 2[sin ...] * cos(x^7 + 3x) *   
       (7x^6 + 3)

You have to know what I have left out, and you must know how to put it 
in. I suggest you complete the derivative of that derivative just 
above, inserting all the expressions that would take the place of the 
...s, then try the problems you're interested in. (All of us at Dr. 
Math had to practice these too.)

- Doctor Santu, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Trigonometry

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