The derivative of 2^{x} is equal to 2^{x}ln2 where ln 2 is the natural logarithm of 2, that is, ln 2 = log_{e}2. In this post, we will find the derivative of 2^{x} by the first principle of derivatives.

## Derivative of 2^{x} from First Principle

We know that the derivative of a function f(x) by the first principle is given by the limit below:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

In this formula, we put f(x)=2^{x}. Then by the first principle, the derivative of 2^{x} is given by

$\dfrac{d}{dx}(2^x)=\lim\limits_{h \to 0} \dfrac{2^{x+h}-2^x}{h}$

$=\lim\limits_{h \to 0}\dfrac{2^x \cdot 2^h -2^x}{h}$, here we have used the indices rule: $a^{m+n}=a^m\cdot a^n$

= $\lim\limits_{h \to 0}\dfrac{2^x (2^h -1)}{h}$

= $2^x \lim\limits_{h \to 0}\dfrac{2^h -1}{h}$

= $2^x \log_e 2$ by the limit formula: $\lim\limits_{t \to 0} \dfrac{a^t-1}{t}$ =log_{e}a.

$=2^x \ln 2$

Thus, the derivative of 2x is $2^x\ln 2$ and this is obtained by the first principle of derivatives.

**Note**: In a similar way as above, one can obtain the derivative of a^{x} by the first principle which is equal to a^{x} ln a.

**Also Read:**

**Question:** Find the derivative of 2^{2x}.

*Answer:*

Let z=2x. Then we have $\dfrac{dz}{dx}=2$. Now by the chain rule, the derivative of 2^{2x} is given as follows:

$\dfrac{d}{dx}(2^{2x})$ $=\dfrac{d}{dz}(2^z) \cdot \dfrac{dz}{dx}$

$=2^z \ln 2 \times 2$ as we have from the above that $\dfrac{d}{dx}(2^x)=2^x\ln 2$

= 2^{2x+1} ln 2 as z=2x.

Thus, the derivative of 2^{2x} is equal to 2^{2x+1} ln 2.

## FAQs

**Q1: What is the derivative of 2 ^{x}?**

**Answer:** The derivative of 2^{x} is equal to 2^{x}ln2.