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Hi PF

In my textbook, the Spanish 6th edition of "Calculus", by Robert A. Adams, at Chapter 3.2, Theorem 1 states:

If ##x>0##.

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

All of this is right. My question is: let's put I know nothing but what I've read until this page: could anybody conclude that ##\ln{(x+h)}-\ln{(x)}## is the area pointed out? As far as I know, the starting point is the Definition 6 of natural logarithm:

"For ##x>0##, be ##A_x## the flat area bounded by ##y=1/t##, the ##t## axis, ##t=1## and ##t=x##. Function ##\ln x## is defined this way:

$$\ln x=\begin{cases}{A_x}&\text{if}& x\geq{1}\\-A_x & \text{if}& 0<x<1\end{cases}$$

As shown in Figure 3.9

Greetings

In my textbook, the Spanish 6th edition of "Calculus", by Robert A. Adams, at Chapter 3.2, Theorem 1 states:

If ##x>0##.

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

**PROOF**For ##x>0## and ##h>0##, ##\ln{(x+h)}-\ln{(x)}## is the area of the shaded part (...) Regard the shaded area at Figure 3.10All of this is right. My question is: let's put I know nothing but what I've read until this page: could anybody conclude that ##\ln{(x+h)}-\ln{(x)}## is the area pointed out? As far as I know, the starting point is the Definition 6 of natural logarithm:

"For ##x>0##, be ##A_x## the flat area bounded by ##y=1/t##, the ##t## axis, ##t=1## and ##t=x##. Function ##\ln x## is defined this way:

$$\ln x=\begin{cases}{A_x}&\text{if}& x\geq{1}\\-A_x & \text{if}& 0<x<1\end{cases}$$

As shown in Figure 3.9

Greetings

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