In this lesson, we are going to see what is the derivative of ln x. We know that ln x is a natural logarithmic function. It means "ln" is nothing but "logarithm with base e". i.e., ln = logₑ. The derivative of ln x is 1/x. We can prove this in two methods.

- By using the first principle (definition of derivative)
- By using implicit differentiation

Let us see what is the derivative of ln x along with its proof in two methods and a few solved examples.

1. | What is the Derivative of ln x? |

2. | Derivative of Natural Log by First Principle |

3. | Derivative of ln x by Implicit Differentiation |

4. | nth Derivative of ln x |

5. | FAQs on Derivative of ln x |

## What is the Derivative of ln x?

The** derivative of ln x** is 1/x. i.e., d/dx (ln x) = 1/x. In other words, the derivative of the natural logarithm of x is 1/x. But how to prove this? Before proving the derivative of ln x to be 1/x, let us prove this roughly by using its graph. For this, we graph the function f(x) = ln x first. We know that the derivative of a function at a point is nothing but the slope of the tangent drawn to the graph of the function at that point. We can clearly see that the slope of the tangent drawn

- at x = 1 is 1
- at x = 2 is 1/2
- at x = 3 is 1/3, and so on.

Thus, the **derivative of ln x is 1/x** which is denoted as d/dx (ln x) = 1/x (or) (ln x)' = 1/x.

### ln Derivative Rules

The ln derivative rule says "the derivative of ln x is 1/x". It is mathematically written as follows:

**d/dx (ln x) = 1/x**(or)**(ln x)' = 1/x**

Let us prove this formula with various methods.

## Derivative of Natural Log by First Principle

Let us prove that the derivative of the natural log is d/dx(ln x) = 1/x using the first principle (the definition of the derivative).

### Proof

Let us assume that f(x) = ln x. By the first principle, the derivative of a function f(x) (which is denoted by f'(x)) is given by the limit,

f'(x) = lim_{h→0} [f(x + h) - f(x)] / h

Since f(x) = ln x, we have f(x + h) = ln (x + h).

Substituting these values in the definition of the derivative,

f'(x) = lim_{h→0} [ln (x + h) - ln x] / h

By a property of logarithms, ln m - ln n = ln (m/n). Applying this, we get

f'(x) = lim_{h→0} [ln [(x + h) / x] ] / h

= lim_{h→0} [ln (1 + (h/x))] / h

Let us assume that h/x = t. From this, h = xt.

Also, when h→0, h/x→0, and hence t→0.

Substituting these values in the above limit,

f'(x) = lim_{t→0} [ln (1 + t)] / (xt)

= lim_{t→0} 1/(xt) ln (1 + t)

By another property of logarithm, m ln a = ln a^{m}. Applying this, we get

f'(x) = lim_{t→0} ln (1 + t)^{1/(xt)}

By a property of exponents, a^{mn} = (a^{m})^{n}. Applying this, we get

f'(x) = lim_{t→0} ln [(1 + t)^{1/t}]^{1/x}

Again by applying ln a^{m} = m ln a,

f'(x) = lim_{t→0} (1/x) ln [(1 + t)^{1/t}]

Since 'x' is irrespective of the variable of the limit, we can write (1/x) outside of the limit.

f'(x) = (1/x) lim_{t→0} ln [(1 + t)^{1/t}] = (1/x) ln lim_{t→0} [(1 + t)^{1/t}]

Using one of the formulas of limits, lim_{t→0} [(1 + t)^{1/t}] = e. Therefore,

f'(x) = (1/x) ln e = (1/x) (1) = 1/x.

Hence we proved that the derivative of ln x is 1/x using the definition of the derivative.

## Differentiation of ln x by Implicit Differentiation

Let us prove that the differentiation of ln x gives d/dx(ln x) = 1/x using implicit differentiation.

### Proof

Assume that y = ln x. Converting this into the exponential form, we get e^{y} = x. Now we will take the derivative on both sides of this equation with respect to x. Then we get

d/dx (e^{y}) = d/dx (x)

By using the chain rule,

e^{y} dy/dx = 1

dy/dx = 1/e^{y}

But we have e^{y} = x. Therefore,

dy/dx = 1/x

Thus, we proved the derivative of ln x to be 1/x using implicit differentiation as well.

## n^{th} Derivative of ln x

Let us observe the first few derivatives of ln x in order to derive its n^{th} derivative:

- d/dx (ln x) = 1/x
- d
^{2}/dx^{2}(ln x) = d/dx (1/x) = d/dx(x^{-1}) = -1x^{-2}=-1/x^{2} - d
^{3}/dx^{3}(ln x) = d/dx (-1x^{-2}) = 2x^{-3}= 2/x^{3} - d
^{4}/dx^{4}(ln x) = d/dx (2x^{-3}) = -6x^{-4}= -6/x^{4} - d
^{5}/dx^{5}(ln x) = d/dx (-6x^{-4}) = 24x^{-5}= 24/x^{5} - ...

Here, note that the signs are alternative + and -, and the coefficients are factorials. By using this pattern, it is easy to notice that the n^{th} differentiation of ln x is given by d^{n}/dx^{n} (ln x) = [ (-1)^{n-1} (n-1)! ]/x^{n}.

**Important Notes on Derivative of ln x:**

- The derivative of ln x is 1/x.
- Though both log x and ln x are logarithms, their derivatives are NOT the same. i.e.,

d/dx ( ln x) = 1/x

d/dx (log x) = 1/(x ln 10) - We know that the domain of ln x is x > 0 and thus, d/dx (ln |x|) = 1/x as well.
- Derivative of ln(f(x)) using chain rule is 1/(f(x)) · f'(x).

☛ **Related Topics:**

- Tangent Line Calculator
- Slope
- Derivative Formulas

## FAQs on Derivative of ln x

### What is the Derivative of Natural Log?

The natural logarithm is denoted by "ln". It is nothing but the common logarithm with base "e". The **derivative of the natural log of x** is 1/x. i.e., d/dx (ln x) = 1/x.

### What is the Result of the Differentiation of ln x?

The differentiation of ln x gives 1/x. Mathematically, we can write it as

- d/dx (ln x) = 1/x
- (ln x)' = 1/x

### What is the Difference Between the Differentiation of ln x and log x?

It is a common misconception for one to assume that the derivative of ln x and the derivative of log x are equal. But the fact is that their derivatives are NOT equal. We have

- d/dx (log x) = 1/(x ln 10)
- d/dx (ln x) = 1/x

### What is the Derivative of 1/x?

To find the derivative of 1/x, we can write it as 1/x = x^{-1}. Then by the power rule, its derivative is -1x^{-2} (or) -1/x^{2}.

### How to Prove that the Derivative of ln x is 1/x?

We can prove that the derivative of ln x is 1/x either by using the definition of the derivative (first principle) or by using implicit differentiation. For detailed proof, click on the following:

- Derivative of ln x by First Principle
- Derivative of ln x by Implicit Differentiation

### What is the Formula for Finding the Derivative of ln x?

The formula of finding the derivative of ln x is, d/dx(ln x) = 1/x. It means that the derivative of ln x is 1/x.

### What is n^{th} Derivative of ln x?

The first derivative of ln x is 1/x. The second derivative is -1/x^{2}. Its third derivative is 2/x^{3}. If we continue this process, the n^{th} derivative of ln x is [(-1)^{n-1} (n-1)!]/x^{n}.

### Is the Derivative of ln x the same as the Derivative of log x?

No, the derivative of ln x is NOT the same as the derivative of log x. The derivative of ln x is 1/x whereas the derivative of log x is 1/(x ln 10).

### What is the Derivative of (ln x)/x?

To find the derivative of (ln x)/x, we use the quotient rule. Then we get [x (1/x) - ln x (1)]/x^{2} = [1 - ln x]/x^{2}.

### What is the Derivative of ln 3x Using the Derivative of ln x?

By property of logarithms, ln 3x = ln 3 + ln x. By differentiating both sides with respect to x, we get d/dx (ln 3x) = d/dx (ln 3) + d/dx (ln x). We know that ln 3 is a constant and hence its derivative is 0. Thus, d/dx(ln 3x) = 0 + 1/x = 1/x.