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

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Registered: 1/6/11
Re: Inversing derivative (NOT an inverse derivative!)
Posted: Jan 16, 2013 2:54 AM
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In article <>,
Anonymous <> 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


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