# Derivative of Natural log (ln(x)) with Proofs and Graphs

The natural logarithm, also denoted as ln(x), is the logarithm of x to base e (euler’s number). The derivative of the natural logarithm is equal to one over x, 1/x. We can prove this derivative using limits or implicit differentiation.

In this article, we’ll look at how to derive the natural logarithmic function. We will go over some fundamentals, definitions, formulas, graph comparisons of underived and derived ln(x), proofs, derivation techniques, and some examples.

##### CALCULUS

Relevant for

Learning how to prove the derivative of natural log, ln(x).

See proofs

##### CALCULUS

Relevant for

Learning how to prove the derivative of natural log, ln(x).

See proofs

## Proofs of the Derivative of Natural Logarithm of x

### Proof of the derivative of ln(x) using the first principle

Before learning the proof of the derivative of the natural logarithmic function, you are hereby recommended to learn/review the first principle of limits, Euler’s number, and L’hopital’s rule as prerequisites.

To review, any function can be derived by equating it to the limit of

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$

Suppose we are asked to get the derivative of

$$f(x) = \ln{(x)}$$

we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \ln{(x+h)} – \ln{(x)} }{h}}$$

With this equation, it is still not possible to express the limit due to the denominator h where if zero is substituted, will be undefined. Therefore, we can check if applying some properties of logarithms can be useful.

The division property of logarithm states that the log of a quotient is the difference of the logs. We can observe that the numerator satisfies this condition. Applying this, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \ln{(x+h)} – \ln{(x)} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \frac{ \ln{\left(\frac{x+h}{x} \right)} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \ln{\left(\frac{x+h}{x} \right)} \cdot \frac{1}{h}}$$

Rearranging, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \frac{1}{h} \cdot \ln{\left(\frac{x+h}{x} \right)} }$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} { \frac{1}{h} \cdot \ln{\left(1 + \frac{h}{x} \right)} }$$

We can try to eliminate the denominator h by substituting

$$h = vx$$

where

$$v = \frac{h}{x}$$

which algebraically proves that as h approaches 0, v also approaches 0.

Substituting, we have

$$\frac{d}{dx} f(x) = \lim \limits_{v \to 0} { \frac{1}{vx} \cdot \ln{\left(1 + v \right)} }$$

Re-arranging, we have

$$\frac{d}{dx} f(x) = \lim \limits_{v \to 0} { \frac{1}{x} \cdot \frac{1}{v} \ln{\left(1 + v \right)} }$$

We may now evaluate the limit of $$\frac{1}{x}$$ as v approaches 0

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{1}{v} \ln{\left(1 + v \right)} }$$

By applying the power property of logarithms to our remaining limit, we have

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \left( \ln{\left(1 + v \right)} \right)^{\frac{1}{v}} }$$

As you realize, the natural log we have in the remaining limit now is exactly the mathematical definition of the Euler’s number e.

If

$$(1 + v)^{\frac{1}{v}} = e$$

based on Euler’s number definition, then

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} {\ln{(e)}}$$

Evaluating ln(e), we know that it is equal to one. Hence, we have

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} {(1)}$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot (1)$$

Therefore, the derivative of the natural logarithm in the form of $$\ln{(x)}$$ is:

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

Alternatively, instead of the Euler’s number definition, we may also evaluate the same remaining limit by applying the L’hopital’s rule.

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{1}{v} \ln{\left(1 + v \right)} }$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{\ln{\left(1 + v \right)}}{v} }$$

This remaining limit satisfies the condition $$\frac{0}{0}$$. Evaluating, we have

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{\ln{\left(1 + v \right)}}{v} }$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{ \frac{1}{1+v} }{1} }$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{1}{1+v} }$$

Evaluating by substituting the approaching value of v, we have

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{1}{1+(0)} }$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { \frac{1}{1} }$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot \lim \limits_{v \to 0} { 1 }$$

$$\frac{d}{dx} f(x) = \frac{1}{x} \cdot (1)$$

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

### Proof of the derivative of ln(x) using implicit differentiation

In this proof, you are hereby recommended to learn/review the derivatives of exponential functions and implicit differentiation.

Suppose we have the equation

$$y = \ln{(x)}$$

In general logarithmic form, it is

$$\log_{e}{x} = y$$

And in exponential form, it is

$$e^y = x$$

Implicitly deriving the exponential form in terms of x, we have

$$e^y = x$$

$$\frac{d}{dx} (e^y) = \frac{d}{dx} (x)$$

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

Isolating $$\frac{dy}{dx}$$, we have

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

We recall that in the beginning, $$y = \ln{(x)}$$. Substituting this to the y of our derivative, we have

$$\frac{dy}{dx} = \frac{1}{e^{(\ln{(x)})}}$$

Evaluating, we now have the derivative of $$y = \ln{(x)}$$

$$y’ = \frac{1}{x}$$

## How to derive the Natural Logarithm of x?

The derivative process of the natural logarithmic of x is very easy assuming you have already learned the concepts behind the purpose of the logarithmic function and how we arrived to its derivative formula through its proofs.

### METHOD 1: Derivative of the natural logarithmic of x by directly applying its derivative formula.

Step 1: Analyze if ln(x) is a function in terms of the same variable x. For example, if the right-hand side of the equation is ln(x), then check if it is a function of the same variable x or f(x).

Note: If ln(x) is a function of a different variable such as f(t) or f(y), it will use implicit differentiation which is out of the scope of this article.

Step 2: Then directly apply the derivative formula of the natural logarithmic function

$$\frac{dy}{dx} = \frac{1}{x}$$

If nothing is to be simplified anymore, then that would be the final answer.

### METHOD 2: Derivative of the natural logarithmic of x by using the derivative of general logarithmic function.

Step 1: Express the function as $$f(x) = \log_{e}{x}$$ instead of $$\ln{(x)}$$

Step 2: Take note that the derivative formula for a general logarithmic function is

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

where b is any positive real number except 0 and 1.

Step 3: Get the derivative of $$f(x) = \log_{e}{x}$$ by applying the formula in step 2.

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

$$\frac{d}{dx} (\log_{e}{x}) = \frac{1}{x\ln{(e)}}$$

and by evaluating ln(e), it is just equal to one. Hence,

$$\frac{d}{dx} (\log_{e}{x}) = \frac{1}{x\cdot(1)}$$

$$\frac{d}{dx} (\log_{e}{x}) = \frac{1}{x}$$

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

If nothing is to be simplified anymore, then that would be the final answer.

### METHOD 3: Derivative of the natural logarithmic of any function u by using the basic chain rule formula.

Step 1: Express the function as $$F(x) = \ln{(u)}$$, where u represents any other function embodied by the natural log.

Step 2: Consider ln(u) as the outside function f(u) and u as the inner function g(x) of the composite function F(x). Hence we have

$$f(u) = \ln{(u)}$$

and

$$g(x) = u$$

Step 3: Get the derivative of the outer function $$f(u)$$, which must use the derivative of the natural logarithmic function, in terms of u.

$$\frac{d}{dx} (\ln{(u)}) = \frac{1}{u}$$

Step 4: Get the derivative of the inner function $$g(x) = u$$ in terms of x. Use the appropriate derivative formula to derive it.

$$\frac{d}{dx} (g(x)) = \frac{d}{dx} (u)$$

Step 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $$f(u)$$ by the derivative of inner function $$g(x)$$

$$\frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$$

$$\frac{dy}{dx} = \frac{1}{u} \cdot \frac{d}{dx} (u)$$

Step 6: Substitute u into $$f(u)$$

Step 7: Simplify and apply any function law whenever applicable to finalize the answer.

## Graph of ln(x)x vs. its derivative

Given the function

$$f(x) = \ln{(x)}$$

the graph is illustrated as

And as we know by now, by deriving $$f(x) = \ln{(x)}$$, we get

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

which is illustrated graphically as

Illustrating both graphs in one, we have

Analyzing the differences of these functions through these graphs, you can observe that the original function $$f(x) = \ln{(x)}$$ has a domain of

$$(0,\infty)$$ or $$x | x > 0$$

and exists within the range of

$$(-\infty, \infty)$$ or all real numbers

whereas the derivative $$f'(x) = \frac{1}{x}$$ has a domain of

$$(-\infty,0) \cup (0,\infty)$$ or $$x | x \neq 0$$

and exists within the range of

$$(-\infty,0) \cup (0,\infty)$$ or $$y | y \neq 0$$

## Examples

Below are some examples of using either the first, second, or third method in deriving a natural logarithmic of a variable in terms of that same variable.

### EXAMPLE 1

Derive: $$f(\beta) = \ln{(\beta)}$$

Solution: Using method 1, we have

Step 1: Analyze if the natural logarithm of $$\beta$$ is a function of the same variable $$\beta$$. In this problem, it is. Hence, proceed to step 2.

Step 2: Directly apply the derivative formula of the natural logarithmic function and derive in terms of $$\beta$$. Since no further simplification is needed, the final answer is:

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

### EXAMPLE 2

Derive: $$f(\lambda) = \ln{(\lambda)}$$

Solution: Using method 2, we have

Step 1: Express the function as $$f(\lambda) = \log_{e}{\lambda}$$ instead of $$\ln{(\lambda)}$$

Step 2: Recalling the derivative formula for a general logarithmic function, we have

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

Step 3: Derive $$f(\lambda) = \log_{e}{\lambda}$$ by applying the formula in step 2.

$$\frac{d}{d\lambda} (\log_{e}{\lambda}) = \frac{1}{\lambda\ln{(e)}}$$

$$\frac{d}{d\lambda} (\log_{e}{\lambda}) = \frac{1}{\lambda\cdot(1)}$$

$$\frac{d}{d\lambda} (\log_{e}{\lambda}) = \frac{1}{\lambda}$$

The final answer is:

$$\frac{d}{d\lambda} (\ln{(\lambda)}) = \frac{1}{\lambda}$$

### EXAMPLE 3

Derive: $$F(x) = \ln{(\sin{(x)})}$$

Solution: Using method 3, we have

Step 1: Express the function as $$F(x) = \ln{(u)}$$, where u represents any the function $$\sin{(x)}$$ embodied by the natural log.

Step 2: Consider ln(u) as the outside function f(u) and u as the inner function g(x) of the composite function F(x). Hence we have

$$f(u) = \ln{(u)}$$

and

$$g(x) = u = \sin{(x)}$$

Step 3: Get the derivative of the outer function $$f(u)$$, which must use the derivative of the natural logarithmic function, in terms of u.

$$\frac{d}{dx} (\ln{(u)}) = \frac{1}{u}$$

Step 4: Get the derivative of the inner function $$g(x) = u$$ in terms of x. Use the derivative of trigonometric sine function to derive it since our u is a sine function.

$$\frac{d}{dx} (g(x)) = {d}{dx} (\sin{(x)})$$

$$\frac{d}{dx} (g(x)) = \cos{(x)}$$

Step 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $$f(u)$$ by the derivative of inner function $$g(x)$$

$$\frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$$

$$\frac{dy}{dx} = \left( \frac{1}{u} \right) \cdot (\cos{(x)})$$

Step 6: Substitute u into $$f(u)$$

$$\frac{dy}{dx} = \left( \frac{1}{(\sin{(x)})} \right) \cdot (\cos{(x)})$$

Step 7: Simplify and apply any function law whenever applicable to finalize the answer.

$$\frac{dy}{dx} = \frac{\cos{(x)}}{\sin{(x)}}$$

The final answer is:

$$\frac{d}{dx} (\ln{(\sin{(x)})}) = \cot{(x)}$$  