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Let $f\colon X\to\mathbb{R}$ be a differentiable function.

What is interpretation of the following quantity:

$$h(x_{0}):=\frac{f'(x_{0})}{f(x_{0})}$$

where $x_{0}\in X$.


My own reaserch.

a)

We know that

$$f'(x_{0})=\lim_{h\to 0}\frac{f(x_{0}+h)-f(x_{0})}{h}$$

so

$$f'(x_{0})h\approx f(x_{0}+h)-f(x_{0})$$

If we take $h=1$ we get

$$f'(x_{0})\approx f(x_{0}+1)-f(x_{0})$$

so we have the following interpretation:

If we increase an argument of function $f$ by 1 unit form level $x_{0}$ then the value of function $f$ will change (approximately) by $f'(x_{0})$ units.

This interpretation is used in economics.

b)

We know that

$$h(x_{0})=\lim_{h\to0}\frac{f(x_{0}+h)-f(x_{0})}{hf(x_{0})}$$

so

$$h(x_{0})h\approx \frac{f(x_{0}+h)-f(x_{0})}{f(x_{0})}$$

and if we take $h=1$ we obtain

$$h(x_{0})\approx \frac{f(x_{0}+1)-f(x_{0})}{f(x_{0})}$$

How can we interpret this quantity?

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  • $\begingroup$ It's the logarithmic derivative: d/dx(log[f(x)]) $\endgroup$ Commented Jul 23, 2016 at 15:44

1 Answer 1

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As you explained, we can write $$\frac{f'(x)}{f(x)}=\frac{df(x)}{dx}\frac{1}{f(x)}\approx \frac{(\Delta f/f)}{\Delta x}$$

where $\Delta x$ is a discrete change in $x$ and $\Delta f$ is the resulting change in $f$.

Thus the ratio is the proportional change in $f$ per unit change in $x$. For example, if $f$ gave the quantity demanded at each given price $x$, then the ratio would tell you the approximate proportional change in quantity demanded for each dollar rise in $x$.

In economics the ratio is known as a semi-elasticity: https://en.wikipedia.org/wiki/Elasticity_of_a_function#Semi-elasticity

Note also that the ratio is the derivative of $\ln f$, which is known as the logarithmic derivative: https://en.wikipedia.org/wiki/Logarithmic_derivative

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  • $\begingroup$ So we can say: If we increase an argument of function $f$ by 1 unit form level $x_0$ then the value of function $f$ will change (approximately) by $h(x_0)$ percent? $\endgroup$
    – Leon
    Commented Jul 23, 2016 at 14:44
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    $\begingroup$ Yes, that is the correct interpretation. $\endgroup$
    – smcc
    Commented Jul 23, 2016 at 15:16

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