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In relation to this question, $\ln|x|$ vs.$\ln(x)$? When is the $\ln$ antiderivative marked as an absolute value?

"On the other hand, if $x$ takes negative values then the derivative of $\ln(−x)$ is $x^{−1}$: you can check this by differentiation."

How?

$$ \frac d {dx} \ln(-x) = \frac 1 {-x} \cdot -1 = \frac 1 x$$

Now plug in a negative value for $x$ and you get $\frac 1 {-x}$. But if I do it right away, $\ln(-x) = \ln(x)$ and all is good.

What am I missing? Why does the order of substitution matter?

Thanks.

Community
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badmax
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1 Answers1

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"Now plug in a negative value for $x$ and you get $\dfrac{1}{-x}$."

No! $1/x$ is still $1/x$. If $x = - (\text{something else})$ then you can say $\dfrac{1}{x} = \dfrac{1}{-\text{something else}}$. But the something else is not $x$.

Robert Israel
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    Right, $\operatorname{something else}$ is not $x$, $\operatorname{something else}$ must be $-x$ since $x$ is $-\operatorname{something else}$. $1/x$ is $\operatorname{something completely different}$ I guess. – AD - Stop Putin - Jun 20 '14 at 06:42