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Derivatives of general logarithmic functions

Consider a logarithmic function \(f(x) = \log_a x\), where \(a>1\) is a constant. By using the change of base rule, we can write \(f(x)\) in terms of the natural logarithm, and then differentiate it:

\[ f(x) = \log_a x = \dfrac{\log_e x}{\log_e a} \] and so, since \(\log_e a\) is a constant, the derivative is \[ f'(x) = \dfrac{1}{x \, \log_e a}. \]

Example

Differentiate \(f(x) = \log_2 (3-7x)\).

Solution

Since

\[ f(x) = \dfrac{\log_e(3-7x)}{\log_e 2}, \] noting that \(\log_e 2\) is just a constant and using the chain rule, we have \begin{align*} f'(x) &= \dfrac{1}{\log_e 2} \cdot \dfrac{1}{3-7x} \cdot (-7)\\ &= \dfrac{-7}{(3-7x) \, \log_e 2}. \end{align*}

Exercise 10

Consider \(f(x) = \log_a (xy)\) and \(g(x) = \log_a x + \log_a y\), where \(a>1\) and \(x,y\) are positive. As our notation suggests, think of \(x\) as a variable and \(y\) as a constant.

  1. Show that \(f'(x) = g'(x)\).
  2. Show that \(f(1) = g(1)\).
  3. Conclude that \(f(x) = g(x)\) for all positive \(x\).

This gives an alternative proof of one of the logarithm laws.

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