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

 

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

$\dfrac {d}{dx} (\log_a x) = \dfrac{1}{ x \ln a} \qquad (I)$

## Examples with Solutions

### Example 1

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
$\dfrac{dy}{du} = \dfrac{1}{u \; \ln 5}$

Evaluate $\dfrac{du}{dx}$
$\dfrac{du}{dx} = 2 x + 2$

Substitute $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ in $(II)$ and write

$\dfrac{dy}{dx} = \dfrac{1}{u \; \ln 5} \; {2x+2} = \dfrac{2x+2}{ (x^2 + 2x -1) \; \ln 5}$

### Example 2

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.