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Find, from the first principle, the derivative of $\ln x$.

My Attempt: \begin{align}f(x)&=\ln (x)\\ f(x+\Delta x)&= \ln (x+\Delta x)\end{align} Now, by formula: \begin{align}f'(x)&=\lim_{\Delta x\to 0} \dfrac {f(x+ \Delta x)-f(x)}{\Delta x}\\ &=\lim_{\Delta x\to 0} \dfrac {\ln(x+\Delta x) -\ln(x)}{\Delta x}\\ &=\lim_{\Delta x\to 0} \dfrac {\ln\left(\dfrac {x+\Delta x}{x}\right)}{\Delta x}\\ &=\lim_{\Delta x\to 0} \dfrac {1}{\Delta x} \ln \left(1+\dfrac {\Delta x}{x}\right)\end{align}

How do.I proceed further?

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    $\begingroup$ What is the definition of $\ln$ in use? $\endgroup$ – Daniel Fischer May 18 '17 at 11:12
  • $\begingroup$ @DanielFischer, Natural log. $\endgroup$ – pi-π May 18 '17 at 11:13
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    $\begingroup$ That much is clear, but how is it defined? as the inverse of $\exp$? As $\int_1^x \frac{1}{t}\,dt$? Something else? $\endgroup$ – Daniel Fischer May 18 '17 at 11:15
  • $\begingroup$ possible duplicate: math.stackexchange.com/questions/1404805/… $\endgroup$ – thesmallprint May 18 '17 at 11:31
  • $\begingroup$ remember for small small 'x' we have that $ ln(1+x) \to x $ $\endgroup$ – Jose Garcia May 18 '17 at 11:32
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If you can use the fact that $e^{\ln x} = x$ then call

$$ f(x) = \ln x $$

Such that

$$ e^{f(x)} = e^{\ln x} = x $$

and

$$ \frac{de^{f(x)}}{dx} = e^{f(x)}\frac{df(x)}{dx} = 1 = x\frac{d\ln x}{dx} $$

Therefore

$$ \frac{d\ln x}{dx} = \frac{1}{x} $$

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