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### Chain Rule and Product Rule: Differentiating

```
Date: 02/19/98 at 18:18:27
From: Kelly
Subject: Differentiating w/ chain rule and product rule

To differentiate:
x^2 * (2-3x)^7
```

```
Date: 02/19/98 at 21:56:59
From: Doctor Sam
Subject: Re: Differentiating w/ chain rule and product rule

Kelly,

Well, there are a couple of things going on here. One is the Product
Rule and the other is the Chain Rule. They are often used together
(as in your example) but they don't have to be. That may be part of

Let's first look at the Product Rule and not worry about the Chain
Rule.

The Product Rule tells how to find the derivative of the product of
two functions.

(f(x) g(x))' = f'(x)g(x) + f(x)g'(x).

If, for example, f(x) = x^2 -2x and g(x) = 3x^2 + 4x and you need the
derivative of (x^2 -2x)(3x^2 + 4x), you can get it by finding:

(x^2 -2x)'(3x^2 + 4x) + (x^2 -2x)(3x^2 + 4x)' =
(2x - 2) (3x^2 + 4x) + (x^2 -2x)(6x + 4)

okay?

Now lets look at the Chain Rule by itself.  The Chain Rule shows how
to find the derivative of functions that have been combined by
composition,
for example, (2x+7)^2
which is a composition of f(x) = 2x+7 and g(x) = x^2.

To find the derivative of (2x+7)^2 = g(f(x)), the Chain Rule says

"differentiate the outside funtion"
[that's g and g'(x) = 2x]

then "differentiate the inside function"
[that's f and f'(x) = 2]

and finally, "multiply g'(f(x)) times f'(x)"
[that's 2(2x+7) times 2].

In symbols these words are usually written as

g(f(x)]' = g'(f(x))f'(x).

And that looks really confusing. I'm not surprised that you don't have

Of course you did it with a substitution. If we write u = 2x+7 then
we are trying to differentiate (2x+7)^2 which is u^2. This derivative
is just 2u but you have to substitute back to get 2(2x+7). You then
multiply by u' = 2 to get the same answer.

In symbols this is usually written as y = u^2 and u = 2x+7. Instead of
writing y' we write dy/du, and instead of u' we write du/dx.

The original problem was to find the derivative of y = (2x+7)^2 where
y is a function of x. (In order to compute a value of y you need a
value of x.)  This derivative is called dy/dx.

You introduced the helping variable u and got y as a function of u
and u as a function of x. This forms a CHAIN of functions from
x to u to y. In order to compute a value of y from a value of x you
must first figure out what u is and the use it to find y. In order to
find the derivative dy/dx of the complicated function (2x+7)^2 we use
the easier functions that are chained together:

x ------->   u  --------> y

to find dy/dx (which means roughly "how fast y changes as x changes")
you must first figure out how fast u changes when you change x and
then figure out how fast y changes when you change u. The Chain Rule
says, in symbols, that

dy/dx = (dy/du)(du/dx)

You multiply the two derivatives to get the derivative that you want.

Now to your problem that uses both the Product Rule and the Chain
Rule:

To find [ x^2 * (2-3x)^7 ]'  first use the product rule:

[ x^2 * (2-3x)^7 ]' = [x^2]' * [(2-3x)^7]  + [x^2][(2-3x)^7]'

The first derivative we can do directly:  [x^2]' = 2x

The second one needs the Chain Rule.  To find [(2-3x)^7]', let

u = 2-3x and y = u^7.

The Chain Rule says that

dy/dx = (dy/du) (du/dx)

dy/du = 7u^6 = 7(2-3x)^6

du/dx = -3

so dy/dx = 7(2-3x)^6 (-3) or -21(2-3x)^6.

Now put it all together:

[ x^2 * (2-3x)^7 ]' = 2x[(2-3x)^7]  + x^2[-21(2-3x)^6]

I hope that helps!

-Doctor Sam,  The Math Forum
Check out our Web site http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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