# Derivatives Introduction

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In this section ask-math gives you the brief idea about derivatives introduction.
In the earlier sections we have learnt about the slope of a line. When the points A(x1,y1) and B(x2,y2) are on the line then the slope 'm' is given by

slope = m= $\frac{y2-y1}{x2 - x1}$

Essentially, the problem of finding the tangent line at a point boils down to the problem of finding the slope of the tangent line at point P(x,f(x)) You can approximate this slope using a secant line* through the point of tangency and a second point on the curve which is very close to the point P. Let that point be Q(x+ $\triangle x$, f(x + $\triangle x$)). Then the slope of this secant line is given by

m = $\frac{y2-y1}{x2 - x1}$

= $\frac{f(x+\triangle x) -f(x)}{x +\triangle x - x}$

=$\frac{f(x+\triangle x) -f(x)}{\triangle x}$

The denominator of the above equation is change in x and numerator
$\triangle y =f(x+\triangle x) -f(x)$ is the change in y.
We will get accurate slope as the value of $\triangle x$ decreases or we can say that we will get more accurate slope as $\triangle x\rightarrow0$

Definition of Tangent Line with Slope m
If is defined on an open interval containing 'x' and if the limit

$\lim_{\triangle x \rightarrow 0}\frac{\triangle y}{\triangle x}=\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$=m

The slope of the tangent line to the graph of at the point is also called the slope of the graph of f(x) at x.

## Examples on derivatives introduction

Example 1 : To find the slope of the graph of f(x) = 2x -3 at x=2, you can apply the definition of the slope of a tangent line.
Solution :

As x=2 so replace (x+$\triangle x$) by (2 + $\triangle x$)

So f(x+$\triangle x$) = 2(x+$\triangle x$)

So, f(2+$\triangle x$) = 2(2+$\triangle x$)

and f(2) = 2(2)-3

Plug in all these values in the definition of tangent

m = $\lim_{\triangle x \rightarrow 0}\frac{\triangle y}{\triangle x}=\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

$\lim_{\triangle x \rightarrow 0}\frac{f(2+\triangle x)-f(2)}{\triangle x} = \lim_{\triangle x \rightarrow 0}\frac{[2(2 +\triangle x)-3)-[2(2)-3]}{\triangle x}$

= $\lim_{\triangle x \rightarrow 0}\frac{4+2\triangle x -3 -4 +3}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{2\triangle x }{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}2$

slope = m = 2
The slope of f at (x,f(x)) =(2,1) is m = 2.
Note : The graph of a linear function has the same slope at any point. This is not true of nonlinear functions.
Example 2: Example 1 : To find the slope of the graph of f(x) =$x^{2}$ at (1,1) and at (-1,2).
Solution :

m = $\lim_{\triangle x \rightarrow 0}\frac{\triangle y}{\triangle x}=\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

Replace x by (x + $\triangle x$) in f(x)

=$\lim_{\triangle x \rightarrow 0}\frac{(x + \triangle x)^{2}-x^{2}}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{x^{2} +2 x \triangle x +\triangle x^{2} -x^{2}}{\triangle x}$

= $\lim_{\triangle x \rightarrow 0}\frac{2 x \triangle x +\triangle x^{2} }{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{ \triangle x (2x +\triangle x) }{\triangle x}$

= $=\lim_{\triangle x \rightarrow 0}{ (2x +\triangle x) }$

m = 2x

Slope at (1,1) = 2x = 2(1) = 2

Slope at (-1,2) = 2x =2(-1) = - 2