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



Virgil
Posts:
8,833
Registered:
1/6/11


Re: Inversing derivative (NOT an inverse derivative!)
Posted:
Jan 16, 2013 2:54 AM


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 



