Before going to find the derivative of log x, let us recall what is "log". "log" is a common logarithm. i.e., it is a logarithm with base 10. If there is no base written for "log", the default base is 10. i.e., log = log₁₀. We can find the derivative of log x with respect to x in the following methods.

- Using the first principle
- Using implicit differentiation
- Using the derivative of ln x

Here we are not only going to find the derivative of log x (with base 10) but also are going to find the derivative of log x with any base.

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

2. | Derivatives of Logs |

3. | Derivative of log x Proof by First Principle |

4. | Derivative of log x Proof by Implicit Differentiation |

5. | Derivative of log x Proof Using Derivative of ln x |

6. | FAQs on Derivative of log x |

## What is the Derivative of log x?

The **derivative of logₐ x** (log x with base a) is 1/(x ln a). Here, the interesting thing is that we have "ln" in the derivative of "log x". Note that "ln" is called the natural logarithm (or) it is a logarithm with base "e". i.e., ln = logₑ. Further, the derivative of log x is 1/(x ln 10) because the default base of log is 10 if there is no base written.

The derivative of log x (base 10) with respect to x is denoted by d/dx (log x) or (log x)'. Thus,

**d/dx(logₐ x) (or) (logₐ x)' = 1/(x ln a)****d/dx(log x) (or) (log x)' = 1/(x ln 10)**

Since the derivative of log x directly follows from the derivative of logₐ x, it is sufficient to prove the latter one. Let us prove this formula using different methods in the upcoming sections.

## Derivatives of Logs

We are going to discuss the derivatives of logs. i.e., the derivatives of both common and natural logarithms. We have already seen that the derivative of logₐ x is 1 / (x ln a). Here, logₐ x is called as a common logarithm. But we have another type of logarithm called the natural logarithm. It is represented as ln x. It is a logarithm with base "e" and hence it can be written as ln x = log_{e}x. Now, we have

d/dx (logₐ x) = 1 / (x ln a)

Substitute a = e on both sides. Then we get:

d/dx (log_{e} x) = 1 / (x ln e)

By the properties of natural logarithms, ln e = 1. So

d/dx (log_{e} x) = 1 / (x · 1)

Thus, d/dx (log_{e} x) = 1.

Replacing log_{e} x with ln x back, we get d/dx (ln x) = 1/x.

Hence, the **derivatives of logs** are:

- d/dx (logₐ x) = 1 / (x ln a) (this is the derivative of common logarithm)
- d/dx (ln x) = 1/x (this is the derivative of natural logarithm)

## Derivative of log x Proof by First Principle

We will prove that d/dx(logₐ x) = 1/(x ln a) using the first principle (definition of the derivative).

**Proof:**

Let us assume that f(x) = logₐ x.

By first principle, the derivative of a function f(x) (which is denoted by f'(x)) is given by the limit,

f'(x) = limₕ→₀ [f(x + h) - f(x)] / h

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

Substituting these values in the equation of first principle,

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

Using a property of logarithms, logₐ m - logₐ n = logₐ (m/n). By applying this,

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

= lim ₕ→₀ [logₐ (1 + (h/x))] / h

Assume that h/x = t. From this, h = xt.

When h→0, h/x→0 **⇒** t→0.

Then the above limit becomes

f'(x) = limₜ→₀ [logₐ (1 + t)] / (xt)

= limₜ→₀ 1/(xt) logₐ (1 + t)

By using property of logarithm, m logₐ a = logₐ a^{m}. By applying this,

f'(x) = limₜ→₀ logₐ (1 + t)^{1/(xt)}

By using a property of exponents, a^{mn} = (a^{m})^{n}. By applying this,

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

By applying the property logₐ a^{m} = m logₐ a,

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

Here, the variable of the limit is 't'. So we can write (1/x) outside of the limit.

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

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

f'(x) = (1/x) logₐ e

= (1/x) (1/logₑ a) (because 'a' and 'e' are interchanged)

= (1/x) (1/ ln a) (because logₑ = ln)

= 1 / (x ln a)

Thus, we proved that the derivative of logₐ x is 1 / (x ln a) by the first principle.

## Derivative of log x Proof by Implicit Differentiation

We will prove that d/dx(logₐ x) = 1 / (x ln a) using implicit differentiation.

**Proof:**

Assume that y = logₐ x. Converting this into the exponential form would give a^{y} = x. By taking the derivative on both sides with respect to x, we get

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

By using the chain rule,

(a^{y} ln a) dy/dx = 1

dy/dx = 1/(a^{y} ln a)

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

dy/dx = 1 / (x ln a)

Hence we proved the derivative of logₐ x to be 1 / (x ln a) using implicit differentiation.

## Derivative of log x Proof Using Derivative of ln x

Note that the derivative of ln x is 1/x. We can convert log into ln using change of base rule. Let us see how.

**Proof:**

Let us assume that f(x) = logₐ x.

By change of base rule, we can write this as,

f(x) = (logₑ x) / (logₑ a)

We know that logₑ = ln. Thus,

f(x) = (ln x) / (ln a)

Now we will find its derivative.

f'(x) = d/dx [(ln x) / (ln a)]

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

= 1 / (ln a) · (1/x)

= 1 / (x ln a)

Thus, we have proved that the derivative of logₐ x with respect to x is 1/(x ln a).

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

Here are some important points to note about the derivative of log x.

- The derivative of logₐ x is 1/(x ln a).
- The derivative of log x is 1/(x ln 10).
- The derivatives of ln x and log x are NOT same.

d/dx(ln x) = 1/x whereas d/dx (log x) = 1/(x ln 10). - As the domain of logₐ x is x > 0, d/dx (logₐ |x|) = 1/(x ln a).

Also, d/dx(log |x|) = 1/(x ln 10).

☛ **Related Topics:**

Here are some topics that are related to the derivative of logₐ x.

- Log Formulas
- Derivative Formulas
- Limit Formulas
- Calculus Calculator

## FAQs on Derivative of log x

### What is the Derivative of log x Base 10 With Respect to x?

The **derivative of log x** (base 10) is 1/(x ln 10). If the log has a base "a", then its derivative is 1/(x ln a). i.e., d/dx(logₐ x) = 1/(x ln a).

### Is the Derivative of log x Equal to 1/x?

No, the derivative of log x is NOT 1/x. In fact, the derivative of ln x is 1/x. But the derivative of log x is 1/(x ln 10).

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

The first derivative of log x is 1/(x ln 10). The second derivative is -1/(x^{2} ln 10). Its third derivative is 2/(x^{3} ln 10). If we continue this process, the n^{th} derivative of log x is [(-1)^{n-1} (n-1)!]/(x^{n }ln 10).

### What is the Derivative of log x Base a?

The derivative of log x with base a is 1/(x ln a). We can prove this using several methods. For more information, click on the following:

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

### What is the Second Derivative of log x?

The first derivative of log x is 1/(x ln 10). This can be written as x^{-1}/(ln 10). Thus, its second derivative is (-1x^{-2})/(ln 10) (or) -1/(x^{2} ln 10).

### What are the Formulas for Derivatives of Logs?

There are two types of formulas for derivatives of logs. One formula talks about the derivative of a common logarithm whereas the other formula talks about the derivative of the natural logarithm.

- For common log: d/dx (logₐ x) = 1 / (x ln a)
- For natural log: d/dx (ln x) = 1/x

### How to Find the Derivative of log(x + 1)?

We know that the derivative of log x is 1/(x ln 10). Again, by the application of chain rule, the derivative of log(x+1) is 1/(x+1) · d/dx(x+1) = 1/(x+1).

### What is the Derivative of log x whole square?

The derivative of (log x)^{2} using the chain rule is 2 log x d/dx(log x) = 2 log x [ 1/(x ln 10)] = (2 log x) / (x ln 10).

### What is the Derivative of log x with base 2?

The derivative of log x with base a is 1/(x ln a). Hence, the derivative of log x with base 2 is 1/(x ln 2).

### What is the Derivative of log x^2?

We know that the derivative of log x is 1/(x ln 10). By applying chain rule, the derivative of log x^{2} is 1/(x^{2}) · (2x) = 2/x.

I am an expert in calculus and logarithmic functions with a deep understanding of derivatives and their applications. My expertise extends to various methods of finding derivatives, including the first principle, implicit differentiation, and the use of logarithmic properties. I have a comprehensive knowledge of the derivative of logarithmic functions, both common and natural.

Now, let's delve into the concepts discussed in the article about finding the derivative of log x:

### 1. Definition of Logarithm

The article begins by defining "log" as a common logarithm with a base of 10, denoted as log₁₀. If no base is specified, the default is 10. The logarithm is then used to introduce the concept of derivatives.

### 2. Derivative of log x

The derivative of logₐ x (log x with base a) is established as 1/(x ln a), and specifically, for log x (base 10), it is 1/(x ln 10). The presence of "ln" indicates the use of the natural logarithm with base "e."

### 3. Derivatives of Logs

The article covers derivatives of both common logarithms (logₐ x) and natural logarithms (ln x). It establishes that the derivative of logₐ x is 1/(x ln a), and for ln x, it is 1/x.

### 4. Proofs of Derivatives

Proofs for the derivative of logₐ x are provided using different methods:
a. **First Principle:** Demonstrated through the definition of derivatives and limits.
b. **Implicit Differentiation:** Utilizing implicit differentiation on the logarithmic equation.
c. **Derivative of ln x:** Transforming logₐ x into ln x using the change of base rule and proving its derivative.

### 5. Important Notes on Derivative of log x

Key points highlighted include:

- The derivative of logₐ x is 1/(x ln a).
- The derivative of log x (base 10) is 1/(x ln 10).
- Emphasizes that the derivatives of ln x and log x are not the same.

### 6. FAQs on Derivative of log x

The article concludes with frequently asked questions, addressing topics such as derivatives of logs, formulas, and specific cases like the second derivative of log x or derivatives with different bases.

This overview showcases a comprehensive understanding of the derivative of logarithmic functions, providing a solid foundation for anyone seeking clarity on this calculus concept.