Derivatives of Exponential Functions
Derivatives of Exponential Functions

The derivative of x2 is equal to 2x. The function x2 denotes the square of x. In this post, we will find the derivative of x square.

Derivative of x2 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 xn is given by $\dfrac{d}{dx}(x^n)$=nxn-1.

Putting n=2 in the above rule, we will get the derivative of x2. Thus, we have that

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

Hence, the derivative of x2 by the power rule of derivatives is 2x.

Next, we will find out the derivative of x2 from first principle, that is, using the limit definition of derivatives.

Derivative of x2 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)=x2

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 = a2+2ab+b2.

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

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

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

= 2x+0

= 2x.

Hence, the derivative of x2 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.