# The Derivative of the Natural Logarithm

The derivative of the natural logarithm is defined the following way:

$f(x) = ln (x)$

$f'(x)=\frac{1}{x}$

The formal proof of the derivative is provided at the bottom of this post.

The following example further explains the derivative of the natural logarithm. Remember that the derivative of a function gives you the slope of that very function, i.e. the first derivative gives you the slope of the function on a given point of evaluation. For instance, assume you have the following two functions:

(1)  $y_{1}=1x$

(2)  $y_{2}=ln(x)$

The first derivative of equation (1) is 1, i.e. $\frac{d y_{1}}{dx}=1$. This means that for each possible $x$ adding an additional $x$ will increase the value of $y_{1}$ by exactly one. You can see this in the graph below. The black line depicts $y_{1}=1x$. In contrary to $y_{1}=1x$ is $y_{2}=ln(x)$ not a constant function. $y_{2}=ln(x)$ is depicted as the red line in the graph below. You can see that the slope of the logarithmic function decreases monotonically in $x$. The slope of $y_{2}= ln(x)$ at a certain $x$ is $\frac{1}{x}$. Next time you do not remember the derivative of the natural logarithm just remember its functional form. You will remember that the slop decreases with an increasing $x$ and remember that the derivative is simply $\frac{1}{x}$.

Formal Derivation of the Derivative of the Natural Logarithm

We begin with

(1)  $y=ln(x)$

we rearrange to (by taking the inverse definition)

(2)  $e^{y}=x$

Take the derivative of both sides with respect to x.

(3)  $e^{y} \frac{dy}{dx}=1$

From (2) we know that $e^{y}=x$. We can substitute $x$ in for $e^{y}$ and get

(4)  $x \frac{dy}{dx}=1$

Rearranging gives us

(5)  $\frac{dy}{dx}=\frac{1}{x}$

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