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Natural logarithm
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== Derivative == The [[derivative]] of the natural logarithm as a [[real-valued function]] on the positive reals is given by<ref name=":1" /> <math display="block">\frac{d}{dx} \ln x = \frac{1}{x}.</math> How to establish this derivative of the natural logarithm depends on how it is defined firsthand. If the natural logarithm is defined as the integral <math display="block">\ln x = \int_1^x \frac{1}{t}\,dt,</math> then the derivative immediately follows from the first part of the [[Fundamental theorem of calculus#First part|fundamental theorem of calculus]]. On the other hand, if the natural logarithm is defined as the inverse of the (natural) exponential function, then the derivative (for {{math|''x'' > 0}}) can be found by using the properties of the logarithm and a definition of the exponential function. From the definition of the number <math>e = \lim_{u\to 0}(1+u)^{1/u},</math> the exponential function can be defined as <math display="block">e^x = \lim_{u\to 0} (1+u)^{x/u} = \lim_{h\to 0}(1 + hx)^{1/h} , </math> where <math>u=hx, h=\frac{u}{x}.</math> The derivative can then be found from first principles. <math display="block">\begin{align} \frac{d}{dx} \ln x &= \lim_{h\to 0} \frac{\ln(x+h) - \ln x}{h} \\ &= \lim_{h\to 0}\left[ \frac{1}{h} \ln\left(\frac{x+h}{x}\right)\right] \\ &= \lim_{h\to 0}\left[ \ln\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}\right]\quad &&\text{all above for logarithmic properties}\\ &= \ln \left[ \lim_{h\to 0}\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}\right]\quad &&\text{for continuity of the logarithm} \\ &= \ln e^{1/x} \quad &&\text{for the definition of } e^x = \lim_{h\to 0}(1 + hx)^{1/h}\\ &= \frac{1}{x} \quad &&\text{for the definition of the ln as inverse function.} \end{align}</math> Also, we have: <math display="block">\frac{d}{dx} \ln ax = \frac{d}{dx} (\ln a + \ln x) = \frac{d}{dx} \ln a +\frac{d}{dx} \ln x = \frac{1}{x}.</math> so, unlike its inverse function <math>e^{ax}</math>, a constant in the function doesn't alter the differential.
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