Derivatives of Logarithmic Functions

Here, we will cover derivatives of logarithmic functions. First, we will derive the equation for a specific case (the natural log, where the base is $e$), and then we will work to generalize it for any logarithm.

Let us create a variable $y$ such that $y = \ln (x)$.

It should be noted that what we want is the derivative of y, or $\frac{dy}{dx}$.

Next, we will raise both sides to the power of $e$ in an attempt to remove the logarithm from the right hand side:

$e^{y} = x$

Applying the chain rule and the property of exponents we derived earlier, we can take the derivative of both sides:

$\dfrac{dy}{dx} \cdot e^{y} = 1$

This leaves us with the derivative

$\dfrac{dy}{dx} = \dfrac{1}{e^{y}}$

Substituting back our original equation of $x = e^{y}$, we find that 

$\dfrac{d}{dx}\ln(x) = \dfrac{1}{x}$

If we wanted, we could go through that same process again for a generalized base, but it is easier just to use properties of logs and realize that 

$\log_{b}(x) = \dfrac{\ln(x)}{\ln(b)}$

Since $\frac{1}{\ln(b)}$ is a constant, we can take it out of the derivative:

$\dfrac{d}{dx}\log_{b}(x) = \dfrac{1}{\ln(b)} \cdot \dfrac{d}{dx}\ln(x)$,

which leaves us with the generalized form of:

$\dfrac{d}{dx}\log_{b}(x) = \dfrac{1}{x \ln(b)}$

We can use the properties of the logarithm, particularly the natural log, to differentiate more difficult functions, such as products with many terms, quotients of composed functions, or functions with variable or function exponents. We do this by taking the natural logarithm of both sides and re-arranging terms using the following logarithm laws:

• $\log \left(\dfrac{a}{b}\right) = \log (a) - \log (b)$
• $\log(a^{n}) = n \log(a)$
• $\log(a) + \log (b) = \log(ab)$

and then differentiating both sides implicitly, before multiplying through by $y$.