The function “log” usually denotes the common logarithm, that is, it is a logarithm with base 10. If no base is mentioned, then one should consider the default base 10. So we have log x = log_{10}x. Note that the logarithm with base e is called the natural logarithm, denoted by ln. That is, log_{e}x = ln x. In this post, we will find the derivative of log x with respect to x using the following methods:

- By chain rule of derivatives.
- By first principle of derivatives.
- By the method of derivatives of implicit functions.

Table of Contents

## What is the derivative of log 3x?

**Answer:** The derivative of log_{a}(3x) is 1/(x log_{e}a).

That is, the derivative of log 3x with base a is equal to 1/(x ln a). So the derivative of log 3x is 1/(x log_{e}10) if the default base is 10.

The formulae for the derivatives of log 3x with different bases are given in the table below:

Log Functions | Derivative |
---|---|

log_{a} 3x | 1/(x log_{e}a) |

log_{10} 3x | 1/(x log_{e}10) |

log_{e} 3x | 1/x |

**Also Read:**

Derivative of e^{sin x} : The derivative of e^{sin x} is cos x e^{sin x}.Derivative of 1/x : The derivative of 1 by x is -1/x ^{2}. |

## Derivative of log 3x by Chain Rule

The chain rule of differentiation says that if f is a function of u and u is a function of x, then the derivative of f with respect to x is equal to

$\dfrac{df}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}.$

Now, we will find the derivative of log_{a}3x (log of 3x with base a) by the chain rule of derivatives.

Note that f=log_{a} 3x is a function of u=3x. So we take

$u = 3x \quad \Rightarrow \dfrac{du}{dx} = 3.$

By the above chain rule of derivatives, the derivative of log 3x is

$\dfrac{d}{dx}(\log_a 3x)=\dfrac{d}{dx}(\log_a u)$ $=\dfrac{d}{du}(\log_a u) \cdot \dfrac{du}{dx}$

$=\dfrac{1}{u \log_e a} \cdot 3$ as the derivative of log_{a}x is 1/(x log_{e}a).

$=\dfrac{1}{3x \log_e a}\cdot 3$ as u=3x.

$=\dfrac{1}{x \log_e a}$.

So the derivative of log 3x with base a is 1/(x log_{e}a). We achieve this by the chain rule of differentiation. Putting a=e, we get the derivative of natural log of 3x which is d/dx(log_{e} 3x) = 1/x as we know that log_{e}e = 1.

## Derivative of ln 3x

**Question:** What is the Derivative of ln 3x?

*Answer:* The derivative of ln 3x is 1/x.

**Proof:**

Note that ln 3x = log_{e} 3x |

∴ d/dx(ln 3x) = d/dx(log_{e} 3x) |

As we know that d/dx(log_{a} 3x)= 1/(x log_{e} a), we get |

d/dx(log_{e} 3x) = 1/(x log_{e} e) = 1/x as ln e =1. |

## Derivative of log 3x from First Principle

The derivative of a function f(x) by first principle is given by the following limit:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$

Take $f(x)=\log_a 3x$ in the above formula. So the derivative of log of 3x with base a using the first principle is

$\dfrac{d}{dx}(\log_a 3x)$ $=\lim\limits_{h \to 0}\dfrac{\log_a 3(x+h)- \log_a 3x}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \frac{3x+3h}{3x}}{h}$ using the formula of $\log_a x -\log_a y$ $=\log_a \frac{x}{y}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \left(1+\frac{h}{x} \right)}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_a \left(1+\frac{h}{x} \right)}{h/x}$ $\times \dfrac{1}{x}$

$=\dfrac{1}{x}\lim\limits_{h \to 0}\dfrac{\log_a \left(1+\frac{h}{x} \right)}{h/x}$

[Let t=h/x. Then t→0 as h→0]

$=\dfrac{1}{x}\lim\limits_{t \to 0}\dfrac{\log_a \left(1+t \right)}{t}$

$=\dfrac{1}{x} \times \log_e a$ as the limit of log_{a}(1+t) / t is log_{e}a when t→0.

$=\dfrac{1}{x\log_e a}$

Hence the derivative of log_{a} 3x is 1/(x log_{e}a) and this is obtained from the first principle of derivatives.

## Derivative of log 3x by Implicit Differentiation

Prove that d/dx(log_{a}3x) = 1/(x log_{e}a) by the method of differentiation for implicit functions.

**Proof:**

Let y = log_{a}3x. By the properties of logarithms, we have

a^{y }= 3x

Differentiating with respect to x, we get that

a^{y} log_{e}a $\frac{dy}{dx}$ = 3

⇒ 3x log_{e}a $\frac{dy}{dx}$ = 3 as we know $a^{\log_a {3x}}=3x$

⇒ $\frac{dy}{dx}$ = 1/(x log_{e}a).

Thus we have shown that the derivative of log_{a} 3x is 1/(x log_{e}a) by implicit differentiation method.

## FAQs on Derivative of log 3x

**Q1: What is the derivative of log 3x?**

Answer: The derivative of log 3x is 1/x if the base is e.

**Q2: What is the derivative of log 3x with base a?**

Answer: The derivative of log_{a} 3x is 1/(x log_{e}a) if the base is a.

**Q3: What is the derivative of ln 3x?**

Answer: The derivative of ln 3x is 1/x. Note that ln 3x denotes the natural logarithm of 3x, that is, logarithm with base e.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.