Derivative of ln2x: Formula, Proof by First Principle, Chain Rule

The derivative of ln2x is 1/x, where ln denotes the natural logarithm. Here, the ln2x derivative is computed using the first principle and the chain rule of derivatives.

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Note that the derivative of ln2x is denoted by d/dx(ln2x), and its formula is given as follows:

$\dfrac{d}{dx}(\ln 2x)=\dfrac{1}{x}.$

That is, the differentiation of ln2x is given by d/dx(ln2x) = 1/x.

Derivative of ln2x

Let us now find the derivative of ln 2x.

ln2x Derivative

Question: How to find the derivative of ln2x?

Answer:

To find ln2x derivative, we will use the following three rules:

  1. ln(ab) = lna + lnb
  2. $\dfrac{d}{dx}$(constant) = 0
  3. $\dfrac{d}{dx}$(lnx) = 1/x.

By the above rules, the derivative of ln2x is given by

$\dfrac{d}{dx}$ (ln 2x) = $\dfrac{d}{dx}$(ln2 + lnx)

= $\dfrac{d}{dx}(\ln 2) + \dfrac{d}{dx}(\ln x)$, by the addition rule of derivatives

= $0 +\dfrac{1}{x}$ as ln2 is a constant

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

So the derivative of ln2x is equal to 1/x.

Read Also: Derivative of ln3x | Derivative of 1/lnx

Derivative of ln2x by First Principle

The derivative of a function f(x) by first principle is given by the limit formula $\dfrac{d}{dx}(f(x))$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$.

Put f(x) = ln2x.

So the derivative of ln2x by the limit definition is

= limh→0 $\dfrac{\ln 2(x+h)-\ln 2x}{h}$

= limh→0 $\dfrac{\ln \Big(\dfrac{2x+2h}{2x} \Big)}{h}$, using the quotient rule of logarithms.

= limh→0 $\dfrac{\ln \Big(1+\dfrac{h}{x} \Big)}{h}$

Let h/x = t, so that t→0 when h→0.
Also, h=tx.

= limt→0 $\dfrac{\ln \big(1+t \big)}{tx}$

= $\dfrac{1}{x}$ limt→0 $\dfrac{\ln \big(1+t \big)}{t}$

= $\dfrac{1}{x} \times 1$, at the limit of ln(1+x)/x is 1 when x→0.

= 1/x.

So the derivative of ln2x is equal to 1/x, and this is obtained by the first principle of derivatives.

Related Topic: Derivative of natural log of x

Derivative of ln2x by Chain Rule

Applying the chain rule using z=2x (so that dz/dx =2), we have that

$\dfrac{d}{dx}(\ln 2x)$

= $\dfrac{d}{dz}(\ln z) \times \dfrac{dz}{dx}$

= $\dfrac{1}{z} \times 2$

= $\dfrac{2}{2x}$ as z=2x.

= 1/x.

So the derivative of ln2x by the chain rule is equal to 1/x.

Also Read: Derivative of ln(1+x)

FAQs

Q1: What is the derivative of ln2x?

Answer: The derivative of the natural logarithm of a function f(x) is equal to f'(x)/f(x). So the derivative of ln2x is equal to (2x)’/2x = 2/2x = 1/x.

Q2: If y=ln2x, then find dy/dx.

Answer: If y=ln2x, then dy/dx =1/x.

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