To differentiate, peel the layers.
The first thing you get to, function-wise, in the above
example function, is the cubing. It's the outside layer.
So differentiate the cube; this means "put a three out
front, and change the exponent to 2":
f'(x) = 3(sin(x^2 + 1))^2 * (the rest that we haven't
found yet)
Now go inside the cube. The sine comes next.
Differentiating a sine means "change the sine to
a cosine":
f'(x) = 3(sin(x^2 + 1))^2 * cos(x^2 + 1) * (the rest
that we haven't found yet)
Now go inside the sine. The polynomial comes
next. They're easy to differentiate:
f'(x) = 3(sin(x^2 + 1))^2 * cos(x^2 + 1) * (2x + 0)
Since differentiating the "x^2 + 1" got us down
to the "x", we're done. Now we simplify:
f'(x) = 6x(cos(x^2 + 1))(sin(x^2 + 1))^2
And that's how you do it!
Hope this helps.
[By the way, if you want to protect against
your e-mail address being spidered for spam
in the newsgroups, don't enter it correctly
anywhere in your message. Use "spam-
guard", as I have below.]
Eliz. Stapel
(remove *nospam* to reply privately)
For math (mostly algebra) links, go to:
http://home.earthlink.net/~stapel/resource.htm
>I am throroughally confused by the "chain rule" in calculus. What is
>it, and can someone give me an example of using it with a practice
>problem?
You can examine the theory from the text. Here is a specific example:
It is usually applied to functions of functions. For example:
y = sin(x^3) is a function of x^3, namely the sine of that function.
Within that is the function of x, namely x^3.
The general practice is to work from the inside out, but not always,
as you see here. I'll use differentials rather than derivative (means
you can perhaps see more clearly the process in this example, and also
good practice in general ifyou are having difficulties.):
dy = d(sin(x^3)) = cos(x^3)d(x^3), and you have the differential wrt
(x^3). In curtailed form, you have recognised that d(sinX) = cos(X)dX
where X is still a function of x.
Although that is correct, it is not complete, since you can see that
d(x^3) = 3x^2dx
So, all together, you have dy = cos(x^3) * 3x^2 dx
or: dy/dx = 3x^2 * cos(x^3)
John.
Much thanks.
email address as provided is phony, a spam deterrant.
Use nfl...@home.com instead :)
>To differentiate, peel the layers.
>
>The first thing you get to, function-wise, in the above
>example function, is the cubing. It's the outside layer.
>So differentiate the cube; this means "put a three out
>front, and change the exponent to 2":
[snip]
Elizabeth, you have completely saved my butt by explaining the chain
rule in a logical, uncomplicated way. Whenever I see a function like
y=[(x-4)^4 + 6x]^8 a voice goes off in my mind. It says "peel the
onion".
Thank you so much.
Glad to help! :-)
Just my two cents.
Gerald Chappell wrote in message <7bfs1e$77r4@indo-news>...
>I don't mean to criticise, but the way I saw it you did not answer the
>original question of the theory of what and why the chain rule operates,
>but instead you merely gave a cookbook layout of how to differentiate a
>compound function.
I don't mean to criticise, but the way I saw it you did not answer the
original question of the theory of what and why the chain rule operates,
but instead you merely gave a cookbook layout of how to differentiate a
compound function. I remember having the theory explained to me, doing
examples the long way (all that dy/dx = dy/du * du/dx) then from this
understanding why you could simply "peel the layers." I think there is
considerable danger in merely instructing someone to follow a recipe
instead of teaching them why the bread rises.
Regards,
Gerald Chappell
Of course, but the person here is not starving. He is in the privileged
position of being able to learn and study. The objective is the
understanding. With that further areas can be explored.
> As I said to the original poster, the Chain Rule
>will make perfect sense =after= he learns how to use it.
>In my years of experience, "after" is usually when it
>clicks for the student. And I was just trying to make
>sure that the student got what he needed.
I know an awful lot of people who can answer maths questions, but if you
asked them why, instead of how, they would be lost. If someone
understands why, the how becomes a formality. Such a person is useful as
he can than go beyond the standard and do more than simply answer textbook
questions.
Regards,
Gerald Chappell