# A new derivative

We can define a logarithmic derivative of a function $f(x)$ as $f^\ddagger (x) = \frac{d}{dx} ln \; f(x) = \frac{f'(x)}{f(x)}$

It's easy to see that it takes on a few nicer properties than derivatives typically do when it comes to quotients and composition

$(fg)^\ddagger = f^\ddagger + g^\ddagger$ $(f/g)^\ddagger = f^\ddagger - g^\ddagger$ $(f(g))^\ddagger = f^\ddagger (g) g'$

We can define common derivatives in terms of the logarithmic derivative:

$(x^n)^\ddagger = n \frac{1}{x}$ $(e^x)^\ddagger = 1$ $(cos(x))^\ddagger = -tan(x)$ $(tan(x))^\ddagger = tan(x) + cot(x)$

Can you find a function $f(x)$ such that $f^\ddagger (x) = f(x)$? Note by Levi Walker
1 year, 10 months ago

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Yes.........there are infinitely many functions.......You need to specify the boundary conditions for a unique solution........Otherwise, it is a simple differential equation......

- 1 year, 10 months ago

It's just an exercise in finding a solution, not the solution :)

- 1 year, 10 months ago

Ohh.........yup.....didn't see that...:P

- 1 year, 10 months ago

Would be nice if you changed the notation for the logarithmic derivative... maybe $f_{\text{L}}(x)$ or $Lf(x)$. As Aaghaz has said somewhere else, a solution would come from a family of solutions to the differential equation $y' = y^2$. Have a look here for more information on this stuff.

- 1 year, 10 months ago

I've seen $\ddagger$ used, so I just adopted that notation. I also find it more aesthetically pleasing tbh, and we all know I'm a sucker for aesthetics.

- 1 year, 10 months ago