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

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.


d/dx eʸ = d/dx (x)

By using the chain rule,

The formula of chain rule is,

dy/dx = (dy/du) × (du/dx)


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).


 f (x) = ln x


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)


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


Let us,

h/x = t 


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


f'(x) = limₜ→₀ ln (1 + t)⁽¹/ˣᵗ⁾

As we know,

 (aᵐ)ⁿ = aᵐⁿ


f'(x) = limₜ→₀ ln [(1 + t)⁽¹/ᵗ⁾]⁽¹/ˣ⁾

Now again putting  ln aᵐ = m ln a


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.


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.



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



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.

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