The derivative of 2^{x} is equal to 2^{x} ln2. Here, ln denotes the natural logarithm (logarithm with base e). In this post, we will find the derivative of 2 to the power x.

## Derivative of 2^{x} Formula

As we know that the derivative of a^{x} is a^{x}ln a, the formula for the derivative of 2^{x} will be as follows:

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

or

$(2^x)’=2^x \ln 2$.

Here, the prime $’$ denotes the first-order derivative.

## What is the Derivative of 2^{x}?

*Answer:* The derivative of 2^{x} is 2^{x}ln 2.

**Explanation: **

We will use the logarithmic differentiation to find the derivative of 2 raised to x. To do so, let us put

z=2^{x}

Taking natural logarithms $\ln$ of both sides, we obtain that

$\ln z=\ln 2^x$

$\Rightarrow \ln z=x\ln 2$ as we know that ln a^{b} = b ln a

Differentiating both sides with respect to x, we get that

$\dfrac{1}{z} \dfrac{dz}{dx}=\ln 2$

$\Rightarrow \dfrac{dz}{dx}=z\ln 2$

$\Rightarrow \dfrac{dz}{dx}=2^x\ln 2$ as z=2^{x}.

Thus, the derivative of 2^x is 2^{x} ln2.

**Also Read:**

Derivative of 2^{x} at x=0

From the above, we obtain the derivative of 2^{x} which is equal to 2^{x} ln 2. So the derivative of 2 to the power x at x=0 will be equal to

$d/dx[2^x]{x=0}$ $=[2^x \ln 2]{x=0}$

= 2^{0} ln 2

= ln 2 as we know that x^{0}=1 for any non-zeero x.

Thus, the derivative of 2^{x} at x=0 is ln 2.

**Also Read: **

## FAQs

**Q1: If y=2 ^{x} then find dy/dx?**

**Answer:** If y=2^{x}, then dy/dx = 2^{x}ln2.