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I am learning logarithmic differentiation. It goes like this:

First we define a function $$L_0(x)=\log|x|=\int_1^{|x|}\frac{1}{t}dt$$ After studying the positive and negative ranges, we know $$L_0'(x)=\frac{1}{x}$$ for all real $x\ne 0$. Apply the above to a function $f(x)$, we have $$g'(x)=(L_0(f(x)))'=L_0'(f(x))f'(x)=\frac{f'(x)}{f(x)}$$ So $f'(x)=g'(x)f(x)$.

However we made an assumption the moment we introduced $L_0(f(x))$: $f(x)\ne 0$. Is the logarithmic derivative method missing the cases when $f(x)=0$? Or to put it in another way, why do we trust the result that it will work for roots of $f(x)$?

Greg Martin
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Dachuan Huang
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    You are correct that this method gives a formula that is valid (only) on any interval on which $f(x)$ is differentiable and does not vanish. When you say "do we trust the result" at zeros of $f$, what result exactly do you mean? $g(x) = \frac1{f(x)}$ is certainly not defined at the zeros of $f$, so we can't expect any statement with $g(x)$ or $g'(x)$ in it to provide information at those points. – Greg Martin Sep 19 '21 at 00:27
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    My book (Apostol calculus Vol 1 p236) showed an example: calculate $f'(x)$ when $f(x)=x^2\cos x(1 + x^4)^{-7}$. The book defined $g(x)=log|f(x)|$, then used logarithmic differentiation, and got the end result $f'(x)=g'(x)f(x)$. Since $f(x)$ is $0$ when $x=\frac{\pi}{2}$, why are we so confident that the end result $f'(x)$ is correct when $x=\frac{\pi}{2}$? The whole process treated $x=\frac{\pi}{2}$ as an undefined point, meaning the whole process did not even consider $x=\frac{\pi}{2}$. – Dachuan Huang Sep 20 '21 at 04:28
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    Or even more simply, if we try this with $f(x)=x^2$, then we get the correct answer $f'(x)=2x$ for all $x\ne0$, and you're asking how we can be certain that the formula is correct when $x=0$ as well (if we didn't know in advance)? That's a solid question.... – Greg Martin Sep 20 '21 at 07:36
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    Yeah that's my question. – Dachuan Huang Sep 21 '21 at 04:43

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Logarithmic differentiation applies wherever the function of interest $f(x)$ is nonzero, and indeed doesn't apply (say anything about $f'(x)$) wherever $f(x)=0.$

For example, logarithmic differentiation per se technically gives the derivative of $\dfrac{e^{-3x}(3x+5)}{7x-1}$ only at $$\mathbb R{\setminus}\left\{-\frac53,\frac17\right\},$$ even though its derivative at $x=-\frac53$ has the same expression.

ryang
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