Derivatives

# Derivative

The operation of calculating the limit of the ratio of the increment of a function to the corresponding increment of an independent variable as the latter tends to zero is called differentiation of the function. The result of operation is called the derivative or the differential coefficient of the function.

## Definition

Let a function $ƒ$ be defined in the interval $(a, b)$. Then, the derivative or the differential coefficient of the function $ƒ$ at a point $x$ of the interval is defined to be the limiting value of $\lim_{\Delta x \to 0} \frac{ƒ(x+\Delta x)-ƒ(x)}{\Delta x}$

Again if $a$ is a fixed point, then, the derivative of $ƒ(x)$ at $x=a$ denoted by $ƒ(a)$ is defined by $\lim_{x \to a} \frac{ƒ(x)-ƒ(a)}{x-a}$ provided that the limit exists. The symbols used to denote the derivative of $ƒ$ with respect to x are $ƒ'(x)$, $\frac{dƒ(x)}{dx}$, $y’$, $\frac{dy}{dx}$.

## Derivative of $x^n$

Now let us calculate the derivatives of the functions $x$, $x^2$ and $x^3$ and deduce the derivative of $x^n$.

1. Let $y=ƒ(x)=x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding small increment in $y$. Then $y+\Delta y=x+\Delta x$ $\Delta y=x+\Delta x-y$ $\Delta y=x+\Delta x-x$ $\Delta y=\Delta x$ $\frac{\Delta y}{\Delta x}=1$ $\therefore \frac{dy}{dx}=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}=1$
2. Let $y=ƒ(x)=x^2$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding small increment in $y$. Then $y+\Delta y=(x+\Delta x)^2$ $\Delta y=(x+\Delta x)^2-y$ $\Delta y=(x+\Delta x)^2-x^2$ $\Delta y=x^2+2x.\Delta x+(\Delta x)^2-x^2$ $\Delta y=2x.\Delta x+(\Delta x)^2$ $\frac{\Delta y}{\Delta x}=2x+\Delta x$ $\therefore \frac{dy}{dx}=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$=\lim_{\Delta x \to 0} (2x+\Delta x)$$=2x$
3. Let $y=ƒ(x)=x^3$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding small increment in $y$. Then $y+\Delta y=(x+\Delta x)^3$ $\Delta y=(x+\Delta x)^3-y$ $\Delta y=(x+\Delta x)^3-x^3$ $\Delta y=x^3+3x^2.\Delta x+3x.(\Delta x)^2+(\Delta x)^3-x^3$ $\Delta y=3x^2.\Delta x+3x.(\Delta x)^2+(\Delta x)^3$ $\frac{\Delta x}{\Delta y}=3x^2+3x.\Delta x+(\Delta x)^2$ $\therefore \frac{dy}{dx}=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$ $=\lim_{\Delta x\to 0}[3x^2+3x.\Delta x+(\Delta x)^2$ $=3x^2$

Therefore, $\text{derivative of }x=\frac{dx}{dx}=x^{1-1}=1$ $\text{derivative of }x^2=\frac{dx^2}{dx}=2.x^{2-1}=2x$ $\text{derivative of }x^3=\frac{dx^3}{dx}=3.x^{3-1}=3x^2$ From these results, we can conclude that the derivative of $x^n$ is $\frac{dx^n}{dx}=nx^{n-1}$

This formula holds not only for natural numbers, it holds for any rational number.

### From First Principles (Using Limit Theorem)

Let $y=x^n$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then, $y+\Delta y=(x+\Delta x)^n$ $\Delta y=(x+\Delta x)^n-y$ $\Delta y=(x+\Delta x)^n-x^n$ $\frac{\Delta y}{\Delta x}=\frac{(x+\Delta x)^n-x^n}{\Delta x}$ $\therefore \frac{dy}{dx}=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ $=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^n-x^n}{(x+\Delta x)-x}$ $=\lim_{(x+\Delta x)\to x}\frac{(x+\Delta x)^n-x^n}{(x+\Delta x)-x}$ $=nx^{n-1}$ $\because \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}$

## Derivative of $(ax+b)^n$ from first principles

Let $y=(ax+b)^n$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then, $y+\Delta y=\{a(x+\Delta x)+b\}^n$ $\Delta y=(ax+a\Delta x+b)^n-y$ $\Delta y=(ax+a\Delta x+b)^n-(ax+b)^n$ $\frac{\Delta y}{\Delta x}=\frac{(ax+a\Delta x+b)^n-(ax+b)^n}{\Delta x}$ $\frac{dy}{dx}=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$ $=\lim_{\Delta x \to 0}\frac{(ax+a\Delta x+b)^n-(ax+b)^n}{\Delta x}$ $=\lim_{\Delta x\to 0} \frac{(ax+a\Delta x+b)^n-(ax+b)^n}{(ax+a\Delta x+b)-(ax+b)}.a$ $=\lim_{(ax+a\Delta x+b)\to (ax+b)} \frac{(ax+a\Delta x+b)^n-(ax+b)^n}{(ax+a\Delta x+b)-(ax+b)}.a$ $=n(ax+b)^{n-1}.a \text{ }\left(\because \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\right)$ $=na(ax+b)^{n-1}$

## Derivative of a constant is zero.

Let $y=ƒ(x)=c$, $c$ is a constant. Then, $y+\Delta y=c$ $\Delta y=c-y$ $\Delta y=c-c$ $\Delta y=0$ $\frac{\Delta y}{\Delta x}=0$ $\frac{dy}{dx}=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=0$

## Relation between Continuity and Differentiability

Consider a function $ƒ(x)$ for which $ƒ'(x)$ exists at $x=a$ i.e. $ƒ'(a)$ is finite. Then we can write $\lim_{h \to 0} [ƒ(a+h)-ƒ(a)]$$=\lim_{h\to 0} \left[\frac{ƒ(a+h)-ƒ(a)}{h}.h\right]$ $=\lim_{h\to 0} \frac{ƒ(a+h)-ƒ(a)}{h}.\lim_{h\to 0} h$ $=ƒ'(a).0=0$ $=\lim_{h\to 0}ƒ(a+h)=ƒ(a)$

Hence, $ƒ(x)$ is continuos at $x=a$. From this, we can see that the differentiability of a function at a point implies the continuity of the function at that point. But this converse is not necessarily true. The continuity of a function at a point does not generally imply the differentiability of the function at that point. For this, consider a function $ƒ(x)$ defined as $ƒ(x)=\left\{ \begin{array} x\sin\frac{1}{x} & \text{ for } x≠0 \\ 0 & \text{ for } x=0 \end{array} \right.$ This function is continuos at $x=0$. Now $ƒ'(0)=\lim_{h\to 0} \frac{ƒ(0+h)-ƒ(0)}{h}$ $=\lim_{h\to 0} \frac{ƒ(h)}{h}$ $=\lim_{h\to 0} \frac{h\sin\frac{1}{h}}{h}$ $=\lim_{h\to 0}\sin\frac{1}{h}$ which is not defined. So $ƒ'(0)$ does not exist.

Hence, the continuity of a function at a point is necessary but it is not the sufficient condition for the existence of derivative of the function at that point.