Proof of Derivative of 2x

The derivative of 2^x can be proved using the following methods:

• By using the First Principle of Derivative
• By using Logarithms

Derivative of 2^x by First Principle of Derivative

To prove derivative of 2^x using First Principle of Derivative, we will use basic limits and exponential formulas which are listed below:

• f'(x) = lim_h→0[f(x + h) – f(x)] / h
• lim_h→0(a^h-1)/h=ln a
• a^m x aⁿ=a^m+n

Let’s start the proof for the derivative of 2^x ,assume that f(x) = 2^x.

By first principle, the derivative of a function f(x) is,

f'(x) = lim_h→0[f(x + h) – f(x)] / h {Using 1}

Since f(x) = 2^x, we have f(x + h) = 2^{(x + h)}.

Substituting these values in (1),

f’ (x) = lim_h→0 [2^{(x + h)} – 2^x]/h

⇒ lim_h→0[2^x · 2^h – 2^x]/h {Using 3}

⇒l im_h→0[2^x · (2^h – 1)]/h

⇒ 2^xlim_h→0[(2^h – 1)]/h {Using 2}

⇒ 2^x · ln 2

Therefore, f'(x) = d/dx [2^x] = 2^x·ln2

Derivative of 2^x by Logarithmic Differentiation

To prove derivative of 2^x using Logarithmic Differentiation, we will use basic formulas which are listed below:

• ln a^b = b ln a
• e^{ln a} = a
• dy/dx[e^x]=e^x

Let’s start the proof for the derivative of 2^x ,assume that y = 2^x.

Taking natural log on both sides,

ln y = ln 2^x

⇒ ln y = x ln 2 {Using 1}

Exponentiate on both sides

⇒ e^{(ln y)} = e^{(x ln 2)}

⇒ y = e^{(x ln 2)} {Using 2}

Take the derivative

⇒ y’ = (ln 2) ·(e^{x ln2}) {Using 3}

⇒ y’ = (ln 2 ) · 2^x

Therefore, f'(x) = d/dx [2^x] = 2^x·ln2

Read More,

• Exponential Functions
• Differentiation Formulas
• Differentiation of Exponential Functions

Derivative of 2^x is 2^xln2. The Derivative of 2^x refers to the process of finding change in 2^x function to the independent variable. The method of finding the derivative for 2^x functions is referred to as exponential differentiation.

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