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.