Derivative of 2^x by First Principle

The derivative of 2^x is equal to 2^xln2 where ln 2 is the natural logarithm of 2, that is, ln 2 = log_e2. 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:

${\displaystyle \frac{d}{dx}}(f(x))$ $=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$

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

${\displaystyle \frac{d}{dx}}(2^x)=\underset{h\to 0}{lim}{\displaystyle \frac{2^{x+h}-2^x}{h}}$

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

= $\underset{h\to 0}{lim}{\displaystyle \frac{2^x(2^h-1)}{h}}$

= $2^x\underset{h\to 0}{lim}{\displaystyle \frac{2^h-1}{h}}$

= $2^x{\log }_{e}2$ by the limit formula: $\underset{t\to 0}{lim}{\displaystyle \frac{a^t-1}{t}}$ =log_ea.

$=2^x\ln 2$

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: 

Simplify 1-sin²x

Derivative of xe^x

Derivative of 10^x

Integration of 1/(1+x²)

Derivative of 1/(1+x²)

Question: Find the derivative of 2^2x.

Answer:

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

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

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

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

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

FAQs