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prove that $f(x)=\ln x:=\int_1^x dt/t$ is a differentiable function:

My Try:

$$\lim_{x\to a} \dfrac{\ln x-\ln a}{x-a}$$

$$=\lim_{x\to a} \dfrac{\ln(\dfrac{x}{a})}{x-a}=?$$

now what ?

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3 Answers 3

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If $f$ is continuous on an interval containing $a$, then $$g(x)=\int_a^xf(t)\,dt$$ is a differentiable function, with derivative $g'(x)=f(x)$. This is the Fundamental Theorem of Calculus, as José Carlos Santos has said.

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It follows from the Fundamental Theorem of Calculus and from your definition of $\ln$ that $\ln'(x)=\frac1x$.

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Hint:

First fundamental theorem of integral calculus.

B.t.w., you have a mistake: $\;\ln x=\displaystyle\int_{\color{red}1}^x\frac{\mathrm dt}t $ for all $x>0$.

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