Basic concepts of Derivative

The theory of Differentiation is the backbone of Calculus. With the help of differentiation, we actually determine the rate of changes of the dependent variable with respect to the independent variable. In this section, we will discuss the concept of derivatives. Here we go. 👩

Few Definitions:

The increment of a variable: Let $x$ be a real variable. Suppose its value changes from $x_0$ to $x_1.$
$x_0$:= The initial value of $x.$
$x_1$:= The final value of $x.$
The difference $x_1-x_0$ is called the increment of $x$. We denote it by $\Delta x$ (Delta $x$)
$ \therefore \Delta x=x_1-x_0$
The increment of a function: Let $f(x)$ be a function in one variable $x.$ Note that the value of $f(x)$ will be changed if we change $x$. As the value of $f(x)$ depends on $x,$ we call $f(x)$ a dependent variable and $x$ an independent variable.

What is the Derivative of a function:

Let $x$ be a real variable and let $f(x)$ be a function of $x.$ Assume that we change the value of $x$ to $x+\Delta x$ $(x \ \rightarrow \ x+\Delta x).$ Here the increment is $\Delta x.$ On the other hand, the value of $f(x)$ will also change to $f(x+\Delta x).$ So the increment of $f(x)$ is equal to $f(x+\Delta x)$ $-f(x).$ We denote this increment by $\Delta f(x)$. Note that the ratio of the increments of the dependent and independent variable is given by
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$\frac{\Delta f(x)}{\Delta x}$ $=\frac{f(x+\Delta x)-f(x)}{\Delta x}.$
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If we tend $\Delta x$ to $0$ $(\Delta x \to 0),$ then the limiting value of $\frac{\Delta f(x)}{\Delta x}$ is called the Derivative or Differential coefficient of $f(x)$ with respect to $x.$ We denote it by $f'(x)$ or $\frac{d}{dx}(f(x)).$
So the derivative of $f(x)$ with respect to $x$ is defined by the following limit:
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$f'(x)=$ $\lim\limits_{\Delta x \to 0} \frac{f(x+\Delta  x)-f(x)}{\Delta x}$
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We write the above discussion as a definition:
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Definition of the Derivative of a function:

Let $y=f(x)$ be a function of  $x.$  Then the derivative of $y$ with respect to $x$ is
$y’=\frac{dy}{dx}$ $=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$
Here $h$ denotes the increment of  $x.$

Some remarks of Derivative: 

  • The method to find the derivative of a function using the above limit definition is called the evaluation of derivatives “from definition” or “from first principle“.

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  • The derivative of $y=f(x)$ at the point $x=a$ is denoted by $f'(a)$ or $\frac{dy}{dx}|_{x=a}.$ Thus, by the definition we have
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    $f'(a)=\frac{dy}{dx}|_{x=a}$ $=\lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$
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    The above limit exists only when both the left-hand limit and the right-hand limit exist. If the left-hand limit exists, that is, the limit
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    $\lim\limits_{h \to 0-} \frac{f(a+h)-f(a)}{h}$
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    exists, we call that limit to be the Left-hand derivative of  $f(x)$ at $x=a$. It is denoted by the symbol $Lf'(a)$ or $f'(a-).$ Similarly, the limit
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    $\lim\limits_{h \to 0+} \frac{f(a+h)-f(a)}{h}$
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    is said to be the Right-hand derivative of $f(x)$ at $x=a,$ andis denoted by $Rf'(a)$ or $f'(a+).$

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  • If $f(x)$ is differentiable at each point on the closed interval $[a,b]$ $(a \leq x \leq b),$ then we say that the function $f(x)$ is differentiable on $[a,b].$

 

Applications of Derivatives:

As it measures the rate of changes, the theory of differentiation is extensively used in many branches of Mathematics, Physics, Chemistry, etc.

 

Solved Examples:

Example 1: From the definition, find $\frac{d}{dx}(x).$

Solution: 

We know that $\frac{d}{dx}(f(x))=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$

Take $f(x)=x$

So by the above definition, we have

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

$=\lim\limits_{h \to 0}\frac{h}{h}$

$=\lim\limits_{h \to 0}1$

$=1$

Therefore, the derivative of $x$ is $1.$

Example 2: From the definition, find $\frac{d}{dx}(x^2).$

Solution: 

One knows that $\frac{d}{dx}(f(x))=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$

Take $f(x)=x^2$

So by the above definition, we have

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

$=\lim\limits_{h \to 0}\frac{x^2+2xh+h^2-x^2}{h}$

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

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

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

$=2x+0$

$=2x$

Therefore, the derivative of $x^2$ is $2x.$

Example 3: From the definition, find $\frac{d}{dx}(\frac{1}{x}).$

Solution: 

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

Take $f(x)=\frac{1}{x}$

So by the above definition, we have

$\frac{d}{dx}(\frac{1}{x})$ $=\lim\limits_{h \to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$

$=\lim\limits_{h \to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}$

$=\lim\limits_{h \to 0}\frac{-h}{xh(x+h)}$

$=\lim\limits_{h \to 0}\frac{-1}{x(x+h)}$

$=-\frac{1}{x(x+0)}$

$=-\frac{1}{x^2}$

Therefore, the derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}.$

Example 4: From the definition, find $\frac{d}{dx}(\sqrt{x}).$

Solution: 

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

Take $f(x)=\sqrt{x}$

So by the above definition, we have

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

$=\lim\limits_{h \to 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim\limits_{h \to 0}\frac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}+\sqrt{x})}$

$[\because (a+b)(a-b)=a^2-b^2]$

$=\lim\limits_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim\limits_{h \to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim\limits_{h \to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}$

$=\frac{1}{\sqrt{x+0}+\sqrt{x}}$

$=\frac{1}{2\sqrt{x}}$

Therefore, the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}.$

 

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