On this page we will see how to calculate the derivative of the natural logarithm using two equivalent forms to express the difference quotient: for \( h \to 0 \) and for \( x \to x_0 \). Formally, as:
\[ \lim_{h \to 0}\frac{\ln(x + h) - \ln(x)}{h}, \quad \lim_{x \to x_0}\frac{\ln(x) - \ln(x_0)}{x - x_0} \]
- Limit of the difference quotient for \( h \to 0 \)
- Limit of the difference quotient for \( x \to x_0 \)
Limit of the difference quotient for \( h \to 0 \)
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 property of logarithms, we can rewrite the numerator in \( ( * ) \) as:
\[ \ln(x + h) - \ln(x) = \ln\left(\frac{x + h}{x}\right) \]
Therefore,
\[ f'(x) = \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x}\right)}{h} \]
To simplify further, we observe that this last expression conceals a remarkable limit. If we let \( t = \frac{h}{x} \), then \( h = x t \). Consequently, when \( h \to 0 \), also \( t \to 0 \). Therefore
\[ f'(x) = \frac{1}{x} \cdot \lim_{t \to 0} \frac{\ln(1 + t)}{t} \stackrel{\text{Remarkable Limit}}{=} \frac{1}{x} \cdot 1 = \frac{1}{x} \]
We find therefore that the derivative of \( \ln(x) \) is
\[ f'(x) = \frac{1}{x}, \quad \forall x > 0 \]
Limit of the difference quotient for \( x \to x_0 \)
Similarly, we calculate the limit when \( x \to x_0 \). Using this definition, the limit of the difference quotient is
\[ f'(x_0) = \lim_{x \to x_0} \frac{\ln(x) - \ln(x_0)}{x - x_0} \]
We exploit the property of logarithms \( \ln( x ) - \ln(x_0) = \ln \left (\frac{x}{x_0} \right ) \). The numerator becomes:
\[ \ln(x) - \ln(x_0) = \ln\left(\frac{x}{x_0}\right) \]
Therefore,
\[ f'(x_0) = \lim_{x \to x_0} \frac{\ln\left(\frac{x}{x_0}\right)}{x - x_0} \qquad (*) \]
To simplify, we let \( u = x - x_0 \), implying that \( x = x_0 + u \). When \( x \to x_0 \), also \( u \to 0 \).
Substituting \( x = x_0 + u \) in the limit \( (*) \), we have
\[ f'(x_0) = \lim_{u \to 0} \frac{\ln\left(\frac{x_0 + u}{x_0}\right)}{u} \]
The argument of the logarithm can be rewritten so as to more easily identify the remarkable limit that will allow us to calculate the derivative we are seeking.
\[ f'(x_0) = \lim_{u \to 0} \frac{\ln\left(1 + \frac{u}{x_0}\right)}{u} \]
If we let \( t = \frac{u}{x_0}\), then \( u = x_0 t \). Moreover \( u \to 0 \) implies \( t \to 0 \):
\[ f'(x_0) = \frac{1}{x_0} \cdot \lim_{t \to 0} \frac{\ln(1+t)}{t} \stackrel{\text{Remarkable Limit}}{=} \frac{1}{x_0} \cdot 1 = \frac{1}{x_0} \]
We conclude that, as in the previous case, the derivative of \( \ln(x) \) is:
\[ f'(x) = \frac{1}{x}, \quad \forall x > 0 \]