# Natural logarithmic derivative trick

Hi chaps and chapesses,

I was wondering if someone could just explain something.

If I have a function which is dependent on $x$, the familiar $f(x)$. Now, if I take the derivative of this, and multiple by $x$ and divide through by $f(x)$.

How does this then become true:

$$\frac{x}{f(x)}\frac{d(f(x))}{dx}=\frac{d\ln{f(x)}}{d\ln{x}}$$

I'm thinking I'm either a.) tired, b.) stupud or c.) both

• "chaps and chapesses"- Interesting? Nov 24, 2015 at 16:54
• It's a British thing. :) Nov 24, 2015 at 16:56
• What does $\frac{df(x)}{d\ln x}$ mean? I mean, $\frac{df}{dx}$ means "the derivative of $f$ in terms of $x$", and since $x$ is a variable in $f$, we know this is defined as $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h},$$ but your expression is different.
– 5xum
Nov 24, 2015 at 17:01

$\frac{x}{f(x)}\frac{d(f(x))}{dx}=\frac{x}{f(x)}\cdot f'(x)=x\cdot \frac{f'(x)}{f(x)}$ $=\large\frac{\frac{f'(x)}{f(x)}}{\frac{1}{x}}=\large\frac{\frac{d}{dx}[\ln(f(x))]}{\frac{d}{dx}[\ln x]}=\frac{d\ln{f(x)}}{d\ln{x}}$
As an alternative to Aniket's answer, we can carefully specify what we mean by $d\ln(x)$ and still get the same result. I'll work from right to left. Let $y(x) = \ln(x)$, whence $x(y) = e^y$. We compute the derivative with respect to $y$ $$\frac{d}{dy} \ln(f(x)) = \frac{dx}{dy} \frac{d}{dx} \ln(f(x)) = \frac{dx}{dy} \frac{1}{f(x)} \frac{d}{dx} f(x).$$ We have that $x(y) = e^y$ so that $\frac{dx}{dy} = e^y = x$. Thus we can write $$\frac{d}{dy} \ln(f(x)) = \frac{dx}{dy}\frac{1}{f(x)}\frac{d}{dx}f(x) = \frac{x}{f(x)}\frac{d}{dx}f(x).$$ This is the meaning of the symbol $\frac{d\ln(f(x))}{d\ln(x)}$.