The “log” function denotes a logarithm with base 10 referred to as common logarithms. So log 2x = log_{10} 2x. If the base is e, then log_{e} 2x = ln 2x is called the natural logarithm. Here, we will find the derivative of log 2x and ln 2x with respect to x. We have:

- The derivative of log 2x is 1/(x log
_{e}10). - The derivative of log
_{a}2x is 1/(x log_{e}a). Here the base is a. - The derivative of ln 2x is 1/x.

Table of Contents

## What is the derivative of log 2x?

We know that the derivative of log_{a}(2x) is 1/(x log_{e}a), that is,

d/dx{log_{a}(2x)} = 1/(x log_{e}a) = 1/(x ln a).

So the derivative of log 2x is 1/(x log_{e}10) where the base is 10.

The formulae for the derivatives of log 2x with different bases are given in the table below:

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

log_{a} 2x | 1/(x log_{e}a) |

log_{10} 2x | 1/(x log_{e}10) |

log_{e} 2x | 1/x |

## Derivative of ln 2x

**Question:** What is the Derivative of ln 2x?

*Answer:* The derivative of ln 2x is 1/x.

**Proof:**

Note that ln 2x = log_{e} 2x |

∴ d/dx(ln 2x) = d/dx(log_{e} 2x) |

As we know that d/dx(log_{a} 2x)= 1/(x log_{e} a), we get |

d/dx(log_{e} 2x) = 1/(x log_{e} e) = 1/x as ln e =1. |

**Also Read:**

Derivative of e : The derivative of e^{sin x}^{sin x} is cos x e^{sin x}.Derivative of ln 3x : The derivative of ln 3x is 1/x. Derivative of sin 3x : The derivative of sin 3x is 3cos 3x. Derivative of 1/x : The derivative of 1 by x is -1/x^{2}. |

## Derivative of log 2x by Chain Rule

If f is a function of u and u is a function of x, then the derivative of f with respect to x by the chain rule is equal to

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

Take f(x) = log_{10} 2x (log of 2x with base 10).

Here u=2x ⇒ du/dx =2.

The derivative of log 2x by the chain rule is

$\dfrac{d}{dx}(\log_{10} 2x)=\dfrac{d}{dx}(\log_{10} u)$ $=\dfrac{d}{du}(\log_{10} u) \cdot \dfrac{du}{dx}$

= $\dfrac{1}{u \log_{e} 10} \cdot 2$ as the derivative of log_{10} x is 1/(x log_{e}10).

= $\dfrac{1}{2x \log_e 10}\cdot 2$ as u=2x.

= $\dfrac{1}{x \log_e 10}$.

So the derivative of log 2x is 1/(x log_{e}10). Now, to obtain the derivative of ln 2x, put a=e. Thus, the derivative of the natural log of 2x is d/dx(log_{e} 2x) = 1/x as we know that log_{e}e = 1.

## Derivative of log 2x from First Principle

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

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

Take f(x)=log_{10} 2x in the above formula. So the derivative of log of 2x with base 10 by the first principle is

$\dfrac{d}{dx}(\log_{10} 2x)$ $=\lim\limits_{h \to 0}\dfrac{\log_{10} 2(x+h)- \log_{10} 2x}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_{10} \frac{2x+2h}{2x}}{h}$. Here, we have used the formula: $\log_a x -\log_a y$ $=\log_a \frac{x}{y}$

$=\lim\limits_{h \to 0}\dfrac{\log_{10} \left(1+\frac{h}{x} \right)}{h}$

$=\lim\limits_{h \to 0}\dfrac{\log_{10} \left(1+\frac{h}{x} \right)}{h/x}$ $\times \dfrac{1}{x}$

$=\dfrac{1}{x}\lim\limits_{h \to 0}\dfrac{\log_{10} \left(1+\frac{h}{x} \right)}{h/x}$

[Let t=h/x. Then t→0 as h→0]

$=\dfrac{1}{x}\lim\limits_{t \to 0}\dfrac{\log_{10} \left(1+t \right)}{t}$

$=\dfrac{1}{x} \times \log_e 10$ as the limit of log_{a}(1+t) / t is log_{e}a when t→0.

$=\dfrac{1}{x\log_e 10}$

Hence the derivative of log_{10} 2x is 1/(x log_{e}10) and this is obtained from the first principle of derivatives.

## Derivative of log 2x by Implicit Differentiation

Prove that d/dx(log_{10}2x) = 1/(x log_{e}10) by the method of differentiation for implicit functions.

**Proof:**

Let y = log_{10}2x. Using the properties of logarithms, we have

10^{y }= 2x

Differentiating with respect to x, we get that

10^{y} log_{e}10 $\frac{dy}{dx}$ = 2

⇒ 2x log_{e}10 $\frac{dy}{dx}$ = 2 as we know $a^{\log_a {2x}}=2x$

⇒ $\frac{dy}{dx}$ = 1/(x log_{e}10).

Thus we have shown that the derivative of log_{10} 2x is 1/(x log_{e}10) by the implicit differentiation method.

## FAQs on Derivative of log 2x

**Q1: What is the derivative of log 2x?**Answer: The derivative of log 2x is 1/x if the base is e.

**Q2: What is the derivative of log 2x with base a?**Answer: The derivative of log_{a} 2x is 1/(x log_{e}a) if the base is a.

**Q3:**

**What is the derivative of ln 2x?**Answer: The derivative of ln 2x is 1/x. Note that ln 2x denotes the natural logarithm of 2x, 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.