The derivative of x^{2} is equal to 2x. The function x^{2} denotes the square of x. In this post, we will find the derivative of x square.

## Derivative of x^{2} by Power Rule

Let us use the power rule of derivatives to find the derivative of x square. The power rule says that the derivative of x^{n} is given by $\dfrac{d}{dx}(x^n)$=nx^{n-1}.

Putting n=2 in the above rule, we will get the derivative of x^{2}. Thus, we have that

$\dfrac{d}{dx}(x^2)$ =2x^{2-1} = 2x^{1} =2x.

Hence, the derivative of x^{2} by the power rule of derivatives is 2x.

Next, we will find out the derivative of x^{2} from first principle, that is, using the limit definition of derivatives.

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

Recall the first principle of derivatives: The derivative of a function f(x) by first principle is given as follows.

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

Let f(x)=x^{2}

So the derivative of x square by the first principle is equal to

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

$=\lim\limits_{h \to 0} \dfrac{x^2+2xh+h^2-x^2}{h}$ using the algebraic identity (a+b)^{2} = a^{2}+2ab+b^{2}.

$=\lim\limits_{h \to 0} \dfrac{2xh+h^2}{h}$

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

$=\lim\limits_{h \to 0} [2x+h]$

= 2x+0

= 2x.

Hence, the derivative of x^{2} by first principle is 2x.

**Question 1:** Find the derivative of $(\ln x)^2$

*Solution: *

Let z=ln x.

So we have $\dfrac{dz}{dx}=\dfrac{1}{x}$.

Then by the chain rule of derivatives, the derivative of \lnx square is equal to

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

$=2z \cdot \dfrac{1}{x}$

$=\dfrac{2}{x} \ln x$ as z=ln x.

Thus, the derivative of (ln x)^{2} is equal to (2/x) ln x.

**Also Read:**

**Derivative of root x + 1 by root x**

## FAQs

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

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