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What is the value of f'(x) at $c$, when $f(x) = \log_x c = e$?

(I understand the answer could be $1/e$) but am unable to substantiate the reasoning. Can someone please help me take the approach?

Thanks.

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    Hmmm -- weird -- the original question, which read "What is the value of$f'(x)$at c, when f(x) = log x$c$= e", seems to suggest the interpretation "What is the value of $f'(x)$ at $c$ when $f(x) = \log x$ and $c = \mathrm e$", to which the answer would indeed be $1/\mathrm e$, and yet you've agreed to a completely different interpretation which, as Michael has pointed out, seems to make no sense -- are you sure you meant $f(x)=\log_x(c)$?2011-12-14

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For the question as currently written: "What is the derivative of $f(x)=\log_x(c)$ at $x=c$?", the simplest thing is to use the change-of-base formula to move $x$ from the base to being the argument of a function.

Since (using $\ln$ for the natural logarithm) $\log_x c = \frac{\ln c}{\ln x}$, differentiating $f(x) = \log_x(c)$ with respect to $x$ is \begin{align*} \frac{d}{dx}\log_x(c) &= \frac{d}{dx}\frac{\ln c}{\ln x}\\ &= \ln(c)\frac{d}{dx}\left(\frac{1}{\ln x}\right)\\ &=\ln(c)\left(\frac{-(\ln x)'}{(\ln x)^2}\right)\\ &= -\frac{\ln(c)}{\ln(x)}\left(\frac{1}{x\ln x}\right)\\ &= -\log_x(c)\left(\frac{1}{x\ln x}\right). \end{align*} Evaluating at $x=c$, we have: f'(c) = -\log_c(c)\left(\frac{1}{c\ln c}\right) = -\frac{1}{c\ln(c)}.

On the other hand, if as joriki surmises, the question was "What is the derivative f'(c) of $f(x)=\log(x)$ [natural logarithm] at $c=e$?" then since f'(x) = \frac{1}{x}, simply plugging $c=e$ yields $\frac{1}{e}$.