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From: Dave L. Renfro on 20 Jun 2007 10:43
dym...(a)gmail.com wrote (in part):

> Can somebody explain how I can evaluate a derivative of
> a function with respect to the natural logarithm of the
> dependent variable? I see this quite a bit in some
> branches of physics (fluid mech/chemists) and it puzzles
> me, i.e. given, y = C * x, find dy/d(ln x).

I think it's just the chain rule. Let y = f(u),
where u = u(x) is a function of x. Then the chain
rule tells us that

dy/dx = (dy/du)*(du/dx).

Hence, dy/du = (dy/dx) / (du/dx).

Letting u = ln(x), this last equality tells you (implicitly)
how to express the derivative of f with respect to ln(x)
in terms of x:

ln(x)-derivative-of-f = (x-derivative-of-f) * 1/x.

I've seen things like this in old (1880's to 1920's)
calculus books, where there will sometimes be exercises
asking for the derivative of tan(x) with respect to cos(x),
the derivative of ln(x) with respect to tanh(x), etc.

Dave L. Renfro

From: Dave L. Renfro on 20 Jun 2007 06:57
Dave L. Renfro wrote (in part):

> ln(x)-derivative-of-f = (x-derivative-of-f) * (1/x)

This should be

ln(x)-derivative-of-f = (x-derivative-of-f) * x

Dave L. Renfro
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