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### Trigonometry and the Chain Rule

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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

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/
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Associated Topics:
High School Calculus
High School Trigonometry

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