Divide first derivative with initial function
Say we have function f(x) and its first derivative f'(x). What do we get when we take f'(x) / f(x) ?
I see this a lot in the mathematics of economic models, but I don't quite get the intuition behind such an operation.
Thanks!
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$\begingroup$The quantity $\frac{f'(x)}{f(x)}\times100$ gives the percentage change for $f(x)$ You can relate it to the usual percentage change calculations, which is similar to $\frac{\Delta N}{N}\times 100$
$\endgroup$ 1 $\begingroup$And, of course, $\dfrac {f^{'}(x)}{f(x)}$ is the drivative of $\log(f(x))$. So in applications in which log scale is an appropriate choice for the response, this expression will arise frequently.
$\endgroup$ $\begingroup$Suppose a quantity $f$ depends on $x$. Let $\%\Delta f$ denote the percent change in $f$ and $\%\Delta x$ denote the percent change in $x$ for a change $\Delta x$. Then\begin{align} \lim_{\Delta x \to 0} \frac{\%\Delta f}{\%\Delta x} &=\lim_{h \to 0} \frac{\frac{f(x + h)}{f(x)} - 1}{\frac{x + h}{x} - 1} \\&= \lim_{h \to 0} \frac{\frac{f(x + h) - f(x)}{f(x)}}{\frac{(x + h) - x}{x}} \\&= \frac{x}{f(x)} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \frac{x f'(x)}{f(x)} \end{align}In other words$$ \frac{f'(x)}{f(x)} \approx \frac{1}{x} \cdot \frac{\% \Delta f}{\% \Delta x} $$for a small $\Delta x$. On the other hand, $x \cdot \%\Delta x = \Delta x$. Thus$$ \frac{f'(x)}{f(x)} \Delta x \approx \%\Delta f $$This says that we can estimate the percent change in $f$, i.e. $\%\Delta f$, by calculating $\frac{f'(x)}{f(x)} \Delta x$. If you don't mind the abuse of language, you might even say that $\frac{f'(x)}{f(x)}$ is the derivative of the percent change in $f$ with respect to $x$. Or, more informally, it's the percent change in $f$ per change in $x$.
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