In this page, we will calculate the derivative of the natural logarithm by using two equivalent forms to express the increment ratio: for \( h \to 0 \) and for \( x \to x_0 \). Formally, we have:
\[ \lim_{h \to 0}\frac{\ln(x + h) - \ln(x)}{h}, \quad \lim_{x \to x_0}\frac{\ln(x) - \ln(x_0)}{x - x_0} \]
Difference Quotient Limit as \( h \to 0 \)
By applying this definition to the function \( \ln(x) \), we obtain:
\[ f'(x) = \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h} \quad ( * ) \]
Using the logarithmic property, we can rewrite the numerator in \( ( * ) \) as:
\[ \ln(x + h) - \ln(x) = \ln\left(\frac{x + h}{x}\right) \]
Thus,
\[ f'(x) = \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x}\right)}{h} \]
To simplify further, observe that this expression hides a notable limit. If we set \( t = \frac{h}{x} \), then \( h = x t \). Consequently, as \( h \to 0 \), we also have \( t \to 0 \). Therefore,
\[ f'(x) = \frac{1}{x} \cdot \lim_{t \to 0} \frac{\ln(1 + t)}{t} \stackrel{\text{Notable Limit}}{=} \frac{1}{x} \cdot 1 = \frac{1}{x} \]
Thus, the derivative of \( \ln(x) \) is:
\[ f'(x) = \frac{1}{x}, \quad \forall x > 0 \]
Difference Quotient Limit as \( x \to x_0 \)
Similarly, we calculate the limit as \( x \to x_0 \). Using this definition, the limit of the increment ratio is:
\[ f'(x_0) = \lim_{x \to x_0} \frac{\ln(x) - \ln(x_0)}{x - x_0} \]
We apply the logarithmic property \( \ln(x) - \ln(x_0) = \ln\left(\frac{x}{x_0}\right) \). The numerator becomes:
\[ \ln(x) - \ln(x_0) = \ln\left(\frac{x}{x_0}\right) \]
Thus,
\[ f'(x_0) = \lim_{x \to x_0} \frac{\ln\left(\frac{x}{x_0}\right)}{x - x_0} \qquad (*) \]
To simplify, let \( u = x - x_0 \), which implies that \( x = x_0 + u \). As \( x \to x_0 \), we also have \( u \to 0 \).
Substituting \( x = x_0 + u \) into the limit \( (*) \), we get:
\[ f'(x_0) = \lim_{u \to 0} \frac{\ln\left(\frac{x_0 + u}{x_0}\right)}{u} \]
The argument of the logarithm can be rewritten to more easily identify the notable limit, which will allow us to compute the derivative we are seeking:
\[ f'(x_0) = \lim_{u \to 0} \frac{\ln\left(1 + \frac{u}{x_0}\right)}{u} \]
If we set \( t = \frac{u}{x_0} \), then \( u = x_0 t \). Moreover, \( u \to 0 \) implies that \( t \to 0 \):
\[ f'(x_0) = \frac{1}{x_0} \cdot \lim_{t \to 0} \frac{\ln(1+t)}{t} \stackrel{\text{Notable Limit}}{=} \frac{1}{x_0} \cdot 1 = \frac{1}{x_0} \]
We conclude that, as in the previous case, the derivative of \( \ln(x) \) is:
\[ f'(x) = \frac{1}{x}, \quad \forall x > 0 \]