5
$\begingroup$

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 $(\ln x)^m$?

1 Answers 1

15

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)

  • 0
    ah cool thx and welcome to http://math.stackexchange.com2012-01-10