# Derivative of 2^x by First Principle

The derivative of 2x is equal to 2xln2 where ln 2 is the natural logarithm of 2, that is, ln 2 = loge2. In this post, we will find the derivative of 2x by the first principle of derivatives.

## Derivative of 2x 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)=2x. Then by the first principle, the derivative of 2x 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: $\lim\limits_{t \to 0} \dfrac{a^t-1}{t}$ =logea.

$=2^x \ln 2$

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 ax by the first principle which is equal to ax ln a.

Simplify 1-sin2x

Derivative of xex

Derivative of 10x

Integration of 1/(1+x2)

Derivative of 1/(1+x2)

Question: Find the derivative of 22x.

Let z=2x. Then we have $\dfrac{dz}{dx}=2$. Now by the chain rule, the derivative of 22x 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 $\dfrac{d}{dx}(2^x)=2^x\ln 2$

= 22x+1 ln 2 as z=2x.

Thus, the derivative of 22x is equal to 22x+1 ln 2.

## FAQs

Q1: What is the derivative of 2x?

Answer: The derivative of 2x is equal to 2xln2.