Derivative of of Logarithm to Any Base : Log_a (x)

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The steps to find the derivative of a logarithmic function to any base are presented.

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 \( \ln \) as
\( y = \log_a x = \dfrac{\ln x}{\ln a} \)

We now evaluate the derivative
\( \dfrac {d}{dx} \left(\log_a x\right) = \dfrac {d}{dx} \left(\dfrac{ \ln x }{\ln a}\right) \)

Noting that \( \ln a \) is a constant and \( \dfrac{d}{dx} (\ln x) = \dfrac{1}{x} \), we obtain

Find the derivatives of
a) \( \quad y = \log_3 (x) \) b) \( \quad y = \log_5 (x^2 + 2x -1) \)

Solution
a) Using the formula in \( (I) \) we obtain
\( \dfrac{dy}{dx} = \dfrac{1}{x \; \ln 3} \)

b) The function in part b) is a composite function of the form \( \quad y = \log_5 u(x) \) with \( u(x) = x^2 + 2x -1 \)
Use the chain rule of differentiation , we write
\( \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \qquad (II) \)

Using the formula in \( (I) \) above
\( \dfrac{dy}{du} = \dfrac{1}{u \; \ln 5} \)

Show that the function \( \quad y = \log_{\frac{1}{2}} (x) \) is decresing in its domain.

Solution
Find the derivative using formula \( (I) \) above
\( \dfrac{dy}{dx} = \dfrac{1}{x \; \ln { \left(\frac{1}{2} \right) }} \)
Note that
\( \ln (\frac{1}{2}) = \ln 1 - \ln 2 = 0 - \ln 2 = - \ln 2 \)
Hence
\( \dfrac{dy}{dx} = \dfrac{1}{ - x \; \ln 2} \)
The domain of the given function \( \quad y = \log_{\frac{1}{2}} (x) \) is the set of all values of \( x \) such that \( x \gt 0 \)
and therefore the derivative \( \dfrac{dy}{dx} = \dfrac{1}{ - x \; \ln 2} \) is negative in the domain of the given function. Since the derivative is negative over domain of the function, the given function \( \quad y = \log_{\frac{1}{2}} (x) \) is a decreasing in its domain.