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.

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.

Let us now find the derivative of ln 2x.

ln2x Derivative

Question: How to find the derivative of ln2x?

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

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 $\dfrac{\ln \Big(1+\dfrac{h}{x} \Big)}{h}$

= 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.