# Derivative of ln x: Proof by Chain and First Principle Rule

Hello, friends in this article you will learn what is derivative of ln x as well as prove the derivative of ln x by the chain rule and first principal rule, and the last also calculate the numerical problem related to derivatives of ln x.

So, without wasting time let's get started.

## What is ln x?

ln x is a natural logarithmic function.  "ln" is a logarithm with base e.

So, we can write "ln" = logₑ.

It is also denoted as logₑˣ this is the inverse function of eˣ.

ln x is always greater than the log x.

## Derivative of ln x

The derivative of ln x is 1/x.

We can derive the derivative of ln x in two ways The first one is by using the chain rule and the second one is using the first principle rule.

### Derivative of ln x Proof by Chain Rule

Let us,

y = ln x

As we know that ln x is the function of exponential e.

So, we can write

x = eʸ

Now we will take the derivative with respect to x on both sides of this equation.

So,

d/dx eʸ = d/dx (x)

By using the chain rule,

The formula of chain rule is,

dy/dx = (dy/du) × (du/dx)

Where,

dy/dx = derivative of y with respect to x

dy/du = derivative of y with respect to u

du/dx = derivative of u with respect to x

After putting these values we can find them,

d/dx eʸ = d/dx (x)

eʸ dy/dx = 1

Shifting eʸ to write side then,

dy/dx = 1/eʸ

But we have x = eʸ Therefore,

dy/dx = 1/x

Thus, we proved the derivative of ln x will be equal to  1/x using the chain rule method.

### Derivative of ln x Proof by First Principle Rule

According to the first principle rule, the derivative limit of a function can be determined by computing the formula:

For a differentiable function y = f (x)

We define its derivative w.r.t  x as :

dy/dx = f ' (x) = limₕ→₀ [f(x+h) - f(x)]/h

f'(x) = limₕ→₀ [f(x+h) - f(x)]/h

This limit is used to represent the instantaneous rate of change of the function f(x).

Let,

f (x) = ln x

So,

f(x + h) = ln (x + h)

Putting these values on the above first principle rules equation.

f'(x) = limₕ→₀ [ln (x + h) - ln x] / h

As we know,

ln (m/n) = ln (m) - ln (n)

So,

f'(x) = limₕ→₀ [ln [(x + h) / x] ] / h

Now dividing the x we will get the value

= lim ₕ→₀ [ln (1 + (h/x))] / h

Now,

Let us,

h/x = t

So,

h = xt

If  h→₀,   so also h/x→₀, and that's why t→₀.

Now putting these values in the above limit we will get,

f'(x) = limₜ→₀ [ln (1 + t)] / (xt)

= limₜ→₀ 1/(xt) ln (1 + t)

As we know,

ln aᵐ = m ln a

So,

f'(x) = limₜ→₀ ln (1 + t)⁽¹/ˣᵗ⁾

As we know,

(aᵐ)ⁿ = aᵐⁿ

So,

f'(x) = limₜ→₀ ln [(1 + t)⁽¹/ᵗ⁾]⁽¹/ˣ⁾

Now again putting  ln aᵐ = m ln a

So,

f'(x) = limₜ→₀ (1/x) ln [(1 + t)⁽¹/ᵗ⁾]

Since the value of x is irrespective of the limit,  we can be taken out 1/x outside of the limit.

So,

f'(x) = (1/x) limₜ→₀ ln [(1 + t)⁽¹/ᵗ⁾]

= (1/x) ln limₜ→₀ [(1 + t)⁽¹/ᵗ⁾]

As we know that one of the formulas of limits is,

limₜ→₀ [(1 + t)⁽¹/ᵗ⁾] = e

So, after putting these value we will get,

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

= (1/x) × (1)

= 1/x.

So here is proved by the first principle derivative rule the derivative of ln x is equal to 1/x.

## Calculation

### Question

Find the derivative of ln (5x - 4)?

### Solution

Let

f(x) = ln (5x- 4)

As we know that the derivative of ln x  is equal to  1/x.

So, through the chain rule formula we will get,

f'(x) = 1/(5x - 4) · d/dx (5x - 4)

= 1/(5x - 4) × (5 - 0)

= 1 / (25x - 20)

= 5/(5x - 4)

So the derivative of the given function will be
5/(5x - 4).

## FAQ Related to Derivative of ln x

### What is ln?

ln is a natural logarithmic function. It is also a logarithm with base e. So, we can write "ln" = logₑ.

### Is the derivative of ln x Always 1/x?

Yes, The derivative of ln x is always equal to 1/x.

### Is the loge the same as ln?

Yes, The loge is the same as ln.

So friends here I discussed all aspects related to the derivative of ln x

I hope you enjoy this topic If you have any doubt then you can ask me through comments or direct mail.

Thank You.