The steps to find the derivative of a logarithmic function to any base are presented below with detailed explanations.
Use of the Change of Base Formula
Let \( y = \log_a x \).
Use the change of base formula to rewrite \( y = \log_a x \) using the natural logarithm \( \ln \):
\[ y = \log_a x = \frac{\ln x}{\ln a} \]We now evaluate the derivative:
\[ \frac{d}{dx} \left(\log_a x\right) = \frac{d}{dx} \left(\frac{\ln x}{\ln a}\right) \]Noting that \( \ln a \) is a constant and \( \frac{d}{dx}(\ln x) = \frac{1}{x} \), we obtain:
Examples with Solutions
Example 1: Basic and Composite Logarithm Derivatives
Problem: Find the derivatives of:
a) \( y = \log_3(x) \)
b) \( y = \log_5(x^2 + 2x - 1) \)
Solution:
a) Using the formula in \( (I) \), we obtain:
\[ \frac{dy}{dx} = \frac{1}{x \ln 3} \]b) The function in part b) is a composite function of the form \( y = \log_5 u(x) \) with \( u(x) = x^2 + 2x - 1 \).
Using the chain rule of differentiation:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \quad (II) \]Using formula \( (I) \):
\[ \frac{dy}{du} = \frac{1}{u \ln 5} \]Evaluate \( \frac{du}{dx} \):
\[ \frac{du}{dx} = 2x + 2 \]Substitute \( \frac{dy}{du} \) and \( \frac{du}{dx} \) into \( (II) \):
\[ \frac{dy}{dx} = \frac{1}{u \ln 5} \cdot (2x + 2) = \frac{2x + 2}{(x^2 + 2x - 1) \ln 5} \]Example 2: Proving a Logarithmic Function is Decreasing
Problem: Show that the function \( y = \log_{\frac{1}{2}}(x) \) is decreasing in its domain.
Solution:
Find the derivative using formula \( (I) \):
\[ \frac{dy}{dx} = \frac{1}{x \ln\left(\frac{1}{2}\right)} \]Note that:
\[ \ln\left(\frac{1}{2}\right) = \ln 1 - \ln 2 = 0 - \ln 2 = -\ln 2 \]Hence:
\[ \frac{dy}{dx} = \frac{1}{-x \ln 2} \]The domain of the given function \( y = \log_{\frac{1}{2}}(x) \) is the set of all values of \( x \) such that \( x > 0 \). Therefore, the derivative \( \frac{dy}{dx} = \frac{1}{-x \ln 2} \) is negative throughout the domain. Since the derivative is negative over the domain of the function, \( y = \log_{\frac{1}{2}}(x) \) is a decreasing function.