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I have a function $f(x)$ and I have plotted it on a log-log plot. I would like to find the location of the inflection point of the function. (note: that means the inflection point in "log-log space" which is not the same as the inflection point of $f(x)$ itself.) I know the following is true: $\frac {d \log y}{d \log x} = \frac{x}{y} \frac{dy}{dx}$. What is the correct formula to find the local maximum of $\frac {d \log y}{d \log x}$?

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  • $\begingroup$ I believe I got the correct answer by taking the derivative of $\frac{x}{y} \frac{dy}{dx}$, and finding the local maximum by setting the result equal to zero and solving for $x$. Is that the correct approach? $\endgroup$ – qdread Jun 15 '18 at 17:53
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Suppose $y=f(x)$ for some function $f$. When you make a log-log plot you are essentially introducing two new variables $$u = \log x, v = \log y$$ and plotting the function $$v=g(u)$$ You can express $g$ in terms of $f$ as follows: $$v = \log y = \log f(x) = \log f(e^u) $$ and therefore the function you are interested in is $g(u) = \log f(e^u)$.

Now, you are looking for the inflection point(s) of $g$. That means you need to study the second derivative of $g$, i.e. $g''(u) = \frac{d^2v}{du^2}$. By the chain rule $$g'(u) = \frac{f'(e^u)e^u}{f(e^u)}$$ and then using the quotient rule and chain rule $$g''(u) =\frac{f(e^u)\left(f''(e^u)e^{2u}+f'(e^u)e^u\right)-\left(f'(e^u)\right)^2e^{2u}}{(f(e^u)^2}$$

In terms of the original variables $x$ and $y$, this is $$\frac{y\left(f''(x)x^{2}+f'(x)x\right)-\left(f'(x)\right)^2x^{2}}{y^2}$$

At an inflection point, the second derivative changes sign, which means that either the quantity above is zero (i.e. its numerator is zero), or the quantity is undefined (i.e. its denominator is zero).

The denominator will be zero if $y=0$, but since you are doing a log-log plot presumably you know that $y>0$ so we can discard this case. The numerator will be zero if $$ y\left(f''(x)x^{2}+f'(x)x\right)=\left(f'(x)\right)^2x^{2}$$ Once again since we are doing a log-log plot I assume you already know that $x>0$, so we can divide through by $x^2$ to get $$ y\left(f''(x)+\frac{f'(x)}{x}\right)=\left(f'(x)\right)^2$$

The criterion $g''(u)=0$ is equivalent to this equation relating $f(x)$ and its derivatives $f'(x)$ and $f(x)$. I don't see any simpler way to write this but perhaps if you know something about $f(x)$ you might be able to simplify it still further.

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