Login [Register]
Don't have an account? Register now to chat, post, use our tools, and much more.
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.

This forum is locked: you cannot post, reply to, or edit 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
Display posts from previous:   
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
    »
» View previous topic :: View next topic  
Page 1 of 1 » All times are GMT - 5 Hours

 

Advertisement