This is an archived, read-only copy of the United-TI subforum , including posts and topic from May 2003 to April 2012. If you would like to discuss any of the topics in this forum, you can visit Cemetech's
Technology & Calculator Open Topic subforum. Some of these topics may also be directly-linked to active Cemetech topics. If you are a Cemetech member with a linked United-TI account, you can link United-TI topics here with your current Cemetech topics.
Math and Science =>
Technology & Calculator Open Topic
| Author |
Message |
|
Flofloflo
Member
Joined: 07 Nov 2007
Posts: 120
|
 Posted: 13 Sep 2009 05:30:45 am Post subject: |
|
|
hello,
can somebody help me with implicit differentiation?
basically, this is my problem:
y=x^(1/n)
x=y^n
d/dx(x) = d/dx(y^n)
d/dx(y^n) = ny^(n-1) * dy/dx
I don't understand that last step. Why do I have to multiply by dy/dx? I am quite unfamiliar with Leibniz notation. My calculus book explains the last step with 'Recalling that y is a function of x, we may evaluate the left-hand side by the chain rule (or the power of a function rule) to get:' and the the thing I wrote down earlier.
Of course, I know they are using the power or chain rule, but why the multiplication?
I am bad with Leibniz notation (I will of course try to improve this, but my math study only just started). I interpret the last formula as:
d/dx(y^n) = dx/dy * dy/dx. Does that make sense?
Help would be appreciated.
-floris
Last edited by Guest on 13 Sep 2009 06:34:39 am; edited 1 time in total |
|
| Back to top |
|
|
thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
|
| Posted: 13 Sep 2009 10:14:12 am Post subject: |
|
|
Remember that y in there is in fact a function y(x). With the chain rule, how do you differentiate y(x)n?
thornahawk |
|
| Back to top |
|
|
Flofloflo
Member
Joined: 07 Nov 2007
Posts: 120
|
| Posted: 13 Sep 2009 11:57:34 am Post subject: |
|
|
Simply n * y(x)^(n-1)
What they do is multiply it by dy/dx. That doesn't make sense to me.... O_o |
|
| Back to top |
|
|
thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
|
| Posted: 13 Sep 2009 07:02:36 pm Post subject: |
|
|
See, n*yn-1 is the derivative with respect to y.That doesn't take into account that y is a function of x.
That's why I was alluding to the chain rule... after taking the derivative of yn with respect to y, you have to multiply the result with the derivative of y with respect to x, because, again, y is a function of x.
Hope that's clear.
thornahawk |
|
| Back to top |
|
|
|
Register to Join the Conversation
Have your own thoughts to add to this or any other topic? Want to ask a question, offer a suggestion, share your own programs and projects, upload a file to the file archives, get help with calculator and computer programming, or simply chat with like-minded coders and tech and calculator enthusiasts via the site-wide AJAX SAX widget? Registration for a free Cemetech account only takes a minute.
»
Go to Registration page
