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I have seen the following proof for the power rule:

$$ \begin{align*} y &= x^n\\ \ln(y) &= n\ln(x)\\ \frac{d}{dx}\ln(y) &= \frac{d}{dx}(n\ln(x))\\ \frac{y'}{y} &= \frac{n}{x}\\ y' &= nx^{n-1} \end{align*} $$

However, what I don't understand is how is this proof valid if $y$ may be negative and then $\ln(y)$ would not be defined. Even if $y'$ is indeed $nx^{n-1}$, isn't this proof a wrong way to achieve the right answer?

Jaideep Khare
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WeakestTopology
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  • One: you can allow complex numbers. Two: If you do not want to allow complex numbers, then you have to proceed more carefully. The "logarithmic derivative" of $f(x)$ is $f'(x)/f(x)$ as long as $f(x)$ is nonzero. No assumption on the sign of $f(x)$. No mention of logarithms. – GEdgar May 10 '17 at 00:36
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    Yes, if $y $ is negative then you can't take the log. Care is needed also with the righthandside. For example $y=\log {(-x)^2} $ is defined and is equal to $y=\log {(x)^2}$ , so you need to take care when working with these quantities. – AnyAD May 10 '17 at 00:40
  • $\ln|y|=n \ln|x|$, etc. – Hans Lundmark May 10 '17 at 06:27
  • Take absolute value before taking logarithm, then apply the result in the addendum here. – ryang Oct 14 '21 at 06:05

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