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Suppose you have the following operator: $$ \left( \frac{d}{dx}\circ x -1 \right), $$ where $\frac{d}{dx}\circ x$ means "multiply by $x$ and then take the derivative of the product". Applying this on $(\ln {x})^m$ gives $$ \left( \frac{d}{dx}\circ x -1 \right)(\ln x)^m= \frac{d}{dx} x (\ln x)^m - (\ln x)^m = (\ln x)^m + m(\ln x)^{m-1} - (\ln x)^m =m(\ln x)^{m-1} $$ This pretty much looks like $\frac{d}{dx}x^m=mx^{m-1}$. So I'm asking if $$ \frac{d}{d \ln{x}} = \left( \frac{d}{dx}\circ x -1 \right) $$ is true in general or just when I apply it on $(\ln x)^m$?

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up vote 13 down vote accepted

Looks like it, because $$ \frac{d f(x)}{d \ln(x)} = \frac{df(e^t)}{dt} = e^t f'(e^t) = xf'(x) = f(x)+xf'(x)-f(x)=\frac{d(xf(x))}{dx} - f(x)$$

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ah cool thx and welcome to math.stackexchange.com –  draks ... Jan 10 '12 at 12:07
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