Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: Inversing derivative (NOT an inverse derivative!)
Replies: 3   Last Post: Jan 16, 2013 7:23 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Virgil

Posts: 9,012
Registered: 1/6/11
Re: Inversing derivative (NOT an inverse derivative!)
Posted: Jan 16, 2013 2:54 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <1852d6e6f1f3cfb205836af0e8bd9dc3@breaka.net>,
Anonymous <noreply@breaka.net> wrote:

> Keeping in mind df(x)/dx = f'(x), if y'(x) = 1/x'(y) and,
> likewise, x'(y) = 1/y'(x), how do you express 1/y'(x)
> in terms of "d"?
> Is it just dx/dy(x)


If y(x) has a unique differentiable local inverse at
the point (x, y(x)), yes

Say z(y) exists such that that z(y(x)) = x in some interval

then by the chain rule z'(y(x))*y'(x) = 1.



>
> d(1/f(x))/dx = -f'(x)/f(x)^2

--





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.