# Derivative of ln x – Natural Logarithm

We’ll look at what the derivative of ln x is in this session. Ln x is a natural logarithmic function, as we know. It signifies that “ln” stands for “logarithm with base e.” ln = loge, for example. There are two ways to find the derivative of ln x.

Making use of the first principle (definition of derivative)

Through the use of implicit differentiation

Let’s look at what the derivative of ln x is, as well as how to prove it using two methods and a few instances.

Derivative Natural Logarithm, ln x:

A number’s natural logarithm is its logarithm to the base e, where e is a mathematical constant roughly equal to 2.718. However, it’s frequently written with the abbreviation ln x, rather than log ex, as you might assume. Meanwhile, in exponential decay problems, logarithms are employed to solve for the half-life, decay constant or unknown duration. They are also used in finance to address problems requiring compound interest and are significant in many disciplines of mathematics and science. However, the differentiation formulas for exponential functions and their inverse functions, known as logarithmic functions, are as follows: Note that the exponential function f( x) = ex has the special property that its derivative is the function itself, f′( x) = ex = f( x).

## Derivative of ln x

1/x is the derivative of ln x. d/dx (ln x) = 1/x, for example. In other words, the natural logarithm of x has a derivative of 1/x. But how do you prove it? However, let us first prove that the derivative of ln x is 1/x in a rough way, utilizing its graph. We start by graphing the function f(x) = ln x. However, the slope of the tangent drawn to the graph of the function at that point is what we call the derivative of a function at a point. The slope of the tangent sketched is readily visible.

at x = 1 is 1

Similarly, at x = 2 is 1/2

Similarly, 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.

## Derivative of ln x Formula

The derivative of ln x with respect to x is

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

(ln x)’ = 1/x

Let us prove this formula in various methods.

## Derivative of ln x by First Principle

Let’s use the first principle to show that d/dx (ln x) = 1/x (the definition of the derivative).

### Derivative of ln x 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,

Then, f'(x) = limₕ→₀ [f(x + h) – f(x)] / h

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

However, substituting these values in the definition of the derivative, we get the following.

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

Similarly, by property of logarithms, ln m – ln n = ln (m/n). Applying this, we get

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

= lim ₕ→₀ [ln (1 + (h/x))] / h

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

Also, when h→0, h/x→0, and hence t→0.

Similarly, by substituting these values for those in the previous limit,

f'(x) = limₜ→₀ [ln (1 + t)] / (xt)

= limₜ→₀ 1/(xt) ln (1 + t)

m ln a = ln am is another logarithm property. When we put this into practise, we obtain

f'(x) = limₜ→₀ ln (1 + t)1/(xt)

Similarly, by a property of exponents, amn = (am)n. Applying this, we get

f'(x) = limₜ→₀ ln [(1 + t)1/t]1/x

Again by applying ln am = m ln a,

f'(x) = limₜ→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.

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

Similarly, using one of the formulas of limits, limₜ→₀ [(1 + t) 1/t] = e.

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

However, using the definition of the derivative, we showed that the derivative of ln x is 1/x.

## Derivative of ln x Natural logarithm by Implicit Differentiation

Let us prove that d/dx (ln x) = 1/x using tacit differentiation.

### Derivative of ln x proof

Assume that y = ln x is the case. We get ey = x when we convert this to exponential form. We’ll now calculate the derivatives of both sides of this equation in terms of x. Then we get

d/dx (ey) = d/dx (x)

By using the chain rule,

ey dy/dx = 1

Similarly, dy/dx = 1/ey

But we have ey = x.

Therefore, dy/dx = 1/x

As a result, we used implicit differentiation to prove that the derivative of ln x is 1/x.

Important Notes about ln x Derivative:

#### Here are some important notes on the 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).

## Solving the Derivative of ln x Natural logarithm

What we’re looking for is the derivative of ln (x). The derivative of ln(x) is 1/x, which is a well-known derivative that most people are familiar with. However, knowing where this formula comes from is always useful, so let’s have a look at the methods for locating this derivative.

Setting y = ln(x) is the first step in calculating the derivative of ln(x) (x). Using the concept of a logarithm, we write y = ln(x) in logarithmic notation. y = log b (x) is identical to b y = x, according to the definition of logarithms. As a result of the logarithm definition and the fact that ln(x) is a logarithm with base e, y = ln(x) is equivalent to ey = x.

Y= ln(x) is equivalent to ey = x

In only a few more steps, we’ll have our formula! After that, we’ll treat y as a function of x and find the derivatives of both sides of the equation with respect to x. On the left side of the equation, we use the chain rule to find the derivative. A two-form rule for finding the derivative of a composition of functions is the chain rule.

## The chain rule:

1. If F(x)= f{g(x)}, then F'(x) = f'{g(x)} × g'(x)
2. Similarly, If y= f(u) and u= g(x), the dy/dx= dx/du × du/dx

However, If f(x) = ex and g(x) = y, then on the left side of the equation, f (g(x)) = ey, where y is a function of x. Because the derivative of e to a variable (such as e x) is the same as the original, the derivative of f'(g(x)) is ey. However, the derivative of ey is ey dy/dx as a result of the chain rule. On the right hand side is x’s derivative, which is 1.

ey = x, take derivative of both side

ey dy/dx = 1

We have (ey) dy/dx = 1 (ey) dy/dx = 1 (ey) dy/dx = Remember that ey = x now. This fact will be used to put x into our equation for ey.

ey dy/dx = 1 plug x in for ey

x dy/dx = 1

As a result, we have the equation (x)dy/dx = 1. Now we’re getting really close! Are you also as giddy as I am about this? We can get dy/dx = 1/x by dividing both sides of this equation by x. Last but not least, remember that y = ln(x) and apply it to our y equation.

x dy/dx = 1, divide by x

dy/dx = 1/x plug in ln x for y

d/dx ln(x) = 1/x

We see the derivative of ln x is 1/x.

However, now that we know that d/dx ln(x) = 1/x, we can see why this ln(x) derivative formula is correct. So, what’s our plan? 1 / x is the derivative of ln(x).

## Derivative of ln(2x) Natural logarithm

Method 1

You use the chain rule :

(f∘g)‘(x)=(f(g(x)))‘=f‘(g(x))⋅g‘(x).

In your case : (f.g)(x)=ln(2x), f(x)=ln(x) and g(x)=2x.

Since f‘(x)=1/x and g‘(x)=2, we have :

(f.g)‘(x)=(ln(2x))‘=1/2x⋅2=1/x.

Method 2

Similarly, the chain rule can be applied here by identifying u=2x and remembering that the chain rule states that

dy/dx=(dy/du)(du/dx)

So, now, for our function ln(u):

dy/du=1/u

And then, for the other part:

du/dx=2

Now, aggregating them:

dy/dx=1/u⋅2=(1/2x)⋅2=1/x

## Derivative of ln(x+1) Natural logarithm

(f∘g)‘(x)=(f(g(x)))‘=f‘(g(x))⋅g‘(x).

In your case : (f∘g)(x)=ln(x+1), f(x)=ln(x) and g(x)=x+1.

Since f‘(x)=1/x and g‘(x)=1, we have :

(f∘g)‘(x)=(ln(x+1))‘=1/x+1⋅1=1/x+1.

## Derivative of ln x square

Applying the chain rule, along with the derivatives d/dx ln(x)=1/x and d/dx(x²)=2x, we have

dy/dx = d/dx ln(x²)

= 1/x² (d/dx x²)

= 1/x² (2x)

=2/x

## Derivative of ln x^3

We can use one of the properties of logs that allow us to write it as:

y = 3 ln x

We can then derive as usual:

y’ = 3 × 1/x

However, we can alternatively utilise the Chain Rule, which involves deriving the log first and multiplying it by the argument’s derivative:

y’ = 1/x³ × 3x²

Simplify,

y = 3/x

## Derivative of ln(3x)

to find out the derivative of ln(3x) then suppose that

ln(3x)=y

e^y=3x

Now use implicit differentiation. Remember that:

dy/dy⋅dy/dx=dy/dx

If you use implicit differentiation…

e^y=3x

Should transform into…

e^y⋅dy/dx=3

Therefore:

dy/dx=3/e^y

Similarly, dy/dx=3/3x

Then, dy/dx=1/x

You could also differentiate it like this

y = ln(3x)

y= ln 3 + ln x

Because of logarithm rules

Hence, dy/dx = 1/x

## Derivative of ln x and log x

d/dx ln x =1/x

This rule can be used to differentiate log x  too.

log x= ln x / ln 10

= d/dx log x =d/dx( lnx / ln10 )

Similarly, =1/ ln 10 × d/dx ln x

Then, =1/ ln 10 × 1x

Similarly, =1/ ln 10^x

Note; here, I’m considering  logx  to be  log10x  only. For any base a, the derivative is 1/ ln a^x.

## Derivative of ln xy

x = y ln (xy)

Or, dx/xy = ln (xy) + [{ y ( dy/dx y + x )} / (xy)}]

Similarly,x dx/dy = x ln (xy) + y dy/dx + x

Therefore, dx/dy (x-y) = x ln (xy) + x

Similarly,dx/dy = x(x+y) / y(x-y)

## Application of Derivative of ln x Natural logarithm

As previously stated, the ln(x) derivative is a well-known derivative that most people can recall. This is due to the fact that this derivative appears frequently in real-world applications. As a result, knowing the derivative is quite handy because you won’t have to go through the process of finding it every time it comes up.

## Natural logarithm

The natural logarithm of a number is its logarithm to the base of the irrational and transcendental number e, which is approximately equal to 2.718281828459. The natural logarithm of x is written as ln x, loge x, or occasionally just log x if the base e is omitted. For clarity, parentheses are occasionally added, resulting in ln(x), loge(x), or log(x) (x). To avoid ambiguity, this is done especially when the argument to the logarithm is not a single symbol.

## Natural logarithm x, ln x

The notations ln x and loge x both unambiguously refer to the natural logarithm of x, while log x without a specific base can also refer to it. This is a common occurrence in mathematics, as well as in various scientific situations and in many computer languages. However, in some cases, such as chemistry, log x can be used to represent the common (base 10) logarithm. In computer science, it can also refer to the binary (base 2) logarithm, especially in the context of time complexity.

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

The process of computing the derivative of x is known as differentiation of x. A modest, very small change in a given function with regard to one of its variables is referred to as differentiation. f'(x) = d[f(x)]/dx is the notation for differentiation of a function f(x). f(x) denotes a function, and dx denotes the variable that the function will be differentiated with. dx/dx, which is equal to 1, can be used to indicate x differentiation. The derivative of the linear function f(x) = ax + b equals a, where a and b are real values. We have a = 1 and b = 0 for f(x) = x. We get the derivative of x equal to 1 using these facts.

### What is the derivative of ln 2 x?

0 is the answer. The natural logarithm function, ln 2, is a constant. If you wish to know the derivative of ln x at x = 2, the answer is 1/2, because the derivative of f(x) = ln x is f'(x) = 1/x, which is f'(2) = 1/2 when evaluated at x = 2.

### Is the derivative of 1 x ln X?

Using the concept of the derivative as a limit, the properties of logarithms, and the definition of e as a limit, we can prove that the derivative of ln(x) is 1/x.

### How do you integrate Ln(x)?

Use Parts Integration as a strategy.

set ln(x) dx, dv = dx, u = ln(x). Then we discover. substitution du = (1/x) dx, v = x. ln(x) dx = u dv ln(x) dx = u dv ln(x) dx

and make use of part-by-part integration = uv – v du. Replace u=ln(x), v=x, and du=(1/x)dx with u=ln(x), v=x, and du=(1/x)dx.

What is the log’s derivative?

The log of a function’s derivative. Logs with a base other than e have a derivative. Let’s start with a graph of the log function with base e, which is expressed as f(x) = loge(x) (or “ln x”).

### What does Ln mean?

The inverse of e is the natural logarithm, which is a fancy phrase for the opposite. The Latin word for logarithmic natural, abbreviated as ln, is somewhat fancy. So, what exactly does inverse or opposite mean? Ex, for example, allows us to plug in time and receive growth. We can plug in growth and get the time it would take using ln(x).

### What are ln and log?

The base 10 logarithm is usually denoted by log(x); it can also be written as log10 (x). The base e logarithm is denoted by ln(x). It can also be spelled loge (x). The natural logarithm of (x) shows you how much you have to raise e to get the number x.

### How do you find the derivative of ln?

The procedure is as follows:

Let’s say y = ln (x). (Logarithm of the natural kind)

Write y = ln(x) in logarithmic form using the definition of a logarithm.

Take the derivative of each side of the equation with respect to x, treating y as a function of x.

To find the derivative, use the chain rule on the left side of the equation.

### How to Calculate the Derivative of x Using Power Rule?

dxn/dx = nxn-1 = nxn-1 is the power rule of differentiation. The derivative of x can be found by substituting n = 1 in this expression. The first principle of derivatives can also be used to evaluate the differentiation of x.

### How do you add LN?

ln(x/y) = ln(x) – ln(y). (Natural logarithm)

ln(x/y) = ln(x) – ln(y).

The natural logarithm of the division of x and y is the difference of the ln of x and ln of y.

Example: ln(7/4) = ln(7) – ln(4)

### How do you differentiate ln3x?

dy/dx = 1/3x x 3 using the chain rule Knowing that ln x equals 1/x, ln 3x equals 1/3x multiplied by the difference of what’s in the bracket. As a result, 1/3x x 3 Equals 3/3x. When the 3s cancel out, you’re left with 1/x.

### What is the derivative of an exponential function?

The differentiation formulas for exponential functions and their inverse functions, known as logarithmic functions, are as follows: It’s worth noting that the exponential function f(x) = e x has the unique property that its derivative is the function itself, f′(x) = e^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 power rule, its derivative is -1x-2 (or) -1/x2.

### What does dy dx mean?

There are a few simple criteria that can be used to quickly differentiate a variety of functions. The derivative of y (with respect to x) is written dy/dx, pronounced “dee y by dee x,” if y = some function of x (in other words, if y is equivalent to an expression including integers and x’s).

• #### What is the derivative of ln(x)?

The derivative of ln(x) is 1/x. Consequently, when differentiating the natural logarithm of x, we obtain 1/x as the result.

• #### How do you find the derivative of ln(x^2)?

To find the derivative of ln(x^2), use the chain rule. First, differentiate ln(u) with respect to u, which is 1/u. Then, differentiate u = x^2 with respect to x, obtaining 2x. Finally, multiply the two results: (1/u) * (2x) = (1/(x^2)) * (2x) = 2/x.

• #### What is the derivative of ln(x+1)?

Applying the chain rule, differentiate ln(u) with respect to u, obtaining 1/u. Next, differentiate u = x + 1 with respect to x, resulting in 1. Multiply the two results: (1/u) * (1) = 1/(x+1).

• #### How can you determine the derivative of ln(x^2+1)?

Utilize the chain rule to find the derivative of ln(x^2+1). Differentiate ln(u) with respect to u, which results in 1/u. Then, differentiate u = x^2 + 1 with respect to x, yielding 2x. Multiply the two results: (1/u) * (2x) = (1/(x^2+1)) * (2x) = 2x/(x^2+1).

• #### What is the derivative of ln(x^3)?

Using the chain rule, differentiate ln(u) with respect to u, obtaining 1/u. Differentiate u = x^3 with respect to x, which results in 3x^2. Multiply the two results: (1/u) * (3x^2) = (1/(x^3)) * (3x^2) = 3x^2/x^3 = 3/x.

• #### Can you explain the proof for the derivative of ln(x)?

To prove the derivative of ln(x), consider the limit definition of a derivative:

d(ln(x))/dx = lim(h->0) [(ln(x+h) – ln(x))/h]

Using the properties of logarithms, rewrite the expression:

= lim(h->0) [ln((x+h)/x)/h] = lim(h->0) [ln(1+h/x)/h]

Now, let u = h/x, so as h -> 0, u -> 0:

= lim(u->0) [ln(1+u)/(ux)]

Applying L’Hôpital’s Rule, differentiate the numerator and denominator with respect to u:

= lim(u->0) [(1/(1+u))/(x)] = 1/x

Thus, the derivative of ln(x) is 1/x.

• #### How do you calculate the derivative of ln(x+y)?

The derivative of ln(x+y) requires implicit differentiation. Differentiate ln(x+y) with respect to x, treating y as a constant:

d(ln(x+y))/dx = (1/(x+y)) * d(x+y)/dx = (1/(x+y)) * 1 = 1/(x+y)