In this page, we will see how to calculate the derivative of the logarithm with base \( b \) using two equivalent forms to express the incremental ratio: for \( h \to 0 \) and for \( x \to x_0 \). Formally:
\[ \lim_{h \to 0}\frac{\log_b(x + h) - \log_b(x)}{h}, \quad \lim_{x \to x_0}\frac{\log_b(x) - \log_b(x_0)}{x - x_0} \]
Table of Contents
Derivative for \( h \to 0 \)
Consider the function \( f(x) = \log_b(x) \). The derivative of \( f(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{\log_b(x + h) - \log_b(x)}{h} \quad ( * ) \]
We use the logarithm base change formula (Properties of Logarithms):
\[ \log_b(x) = \frac{\ln(x)}{\ln(b)} \]
Therefore, the numerator in \( (*) \) becomes:
\[ \log_b(x + h) - \log_b(x) = \frac{\ln(x + h) - \ln(x)}{\ln(b)} \]
Simplifying:
\[ f'(x) = \frac{1}{\ln(b)} \cdot \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h} \]
We already know that:
\[ \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h} = \frac{1}{x} \]
Therefore:
\[ f'(x) = \frac{1}{x \ln(b)} \]
We find that the derivative of \( \log_b(x) \) is:
\[ f'(x) = \frac{1}{x \ln(b)}, \quad \forall x > 0 \]
Derivative for \( x \to x_0 \)
Now, consider the incremental ratio for \( x \to x_0 \):
\[ f'(x_0) = \lim_{x \to x_0} \frac{\log_b(x) - \log_b(x_0)}{x - x_0} \]
Applying the logarithm base change formula (Properties of Logarithms):
\[ \log_b(x) - \log_b(x_0) = \frac{\ln(x) - \ln(x_0)}{\ln(b)} \]
Simplifying:
\[ f'(x_0) = \frac{1}{\ln(b)} \cdot \lim_{x \to x_0} \frac{\ln(x) - \ln(x_0)}{x - x_0} \]
We know that:
\[ \lim_{x \to x_0} \frac{\ln(x) - \ln(x_0)}{x - x_0} = \frac{1}{x_0} \]
Therefore:
\[ f'(x_0) = \frac{1}{x_0 \ln(b)} \]
In this case as well, we obtain:
\[ f'(x) = \frac{1}{x \ln(b)}, \quad \forall x > 0 \]