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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?


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Are you asking Hints for arriving at $f\ '(c)=1/e$?? – Ramana Venkata Dec 14 '11 at 10:48
Is it $f(x)=\log(xc)$ or $f(x)=\log_x (c)$? – lhf Dec 14 '11 at 10:55
the latter is the one correct, @Lhf – Nikhil Mulley Dec 14 '11 at 11:07
yes @RamanaVenkata – Nikhil Mulley Dec 14 '11 at 11:25
What exactly does "$f(x)=\log_x c = e$" mean? Is it supposed to identify the value of $c$? That would depend on what $x$ is. Is it supposed to identify the value of $x$ at which you want to evaluate $f'(x)$? That seems to conflict with the earlier phrase "at $c$", which seems to mean you want to evaluate $f'(x)$ at $x=c$. – Michael Hardy Dec 14 '11 at 11:48

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 $\frac{1}{e}$.

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