Basic Concept of Derivative for Grade 11

In this post you will study about the basic concepts of derivative, its formula, its importance and some solved problem. Derivative is one of the important concept of higher mathematics and its property is used in various other branches like Physics, Mechanics, Statistics etc., so make sure you fully understand the base of this topic.

Understanding Derivative

To understand this concept, let us recall the calculation of slope of a line.

Constant Slope

From the above calculation we can see that the slope of the line is 1.
In this type of lines the slope is constant throughout the graph, hence we don’t have any problems regarding the slope calculation.

Variable Slope

Now there are lines whose value of slope changes throughout the graph. For these lines we calculate the average value of the slope.

From the above figure you can see that at each of the point A, B, C and D the slope is different, which means that the slope is changing throughout the line.

One simple way to deal with this problem is to find the average slope value for the entire line

Average Slope Calculation

From the above illustration you can see that we have selected two points A & D on the line and used their value to find the average slope calculation. In this case the value of average slope is 0.75

Now this method do not solve our purpose when we need accurate measurement of slope.
In this case we will change our calculation methodology, instead of finding slope of entire line, we will try to find slope at a particular point.

But how can we calculate the slope at a point?
Don’t we need atleast two points for the calculation of slope?

For the calculation of slope at a point we will definitely take two point, but this time the other point is very close such that the distance between the point is approaching zero (Remember the concept of Limits !!)

Slope at Exact Point

Let us zoom the above graph so that we can select points which are close enough.

From the above illustration we can say that:
x\ changes\ from\ x\ to\ \Delta x\\ \\ y\ changes\ from\ y\ to\ \Delta y \\\ \\ Slope\ at\ A\ =\frac{( y+\Delta y) -y}{( x+\Delta x) -x}\\\ \\ Slope\ at\ A\ =\ \frac{\Delta y}{\Delta x}\\\ \\ where\ \Delta y\ and\ \Delta x\ are\ so\ small\ that\ they\ are\ considered\ approaching\ zero

This calculation of the slope at point A is also known as the calculation of derivative at the point A.
Note that the value of slope (or derivative) change as we move our position from A to different point.

Method 2 of Derivative Calculation

x\ changes\ from\ x\ to\ \Delta x\\\ \\ y\ changes\ from\ f( x) \ to\ f( x+\Delta x)\\\ \\ we\ know\ that\ \Delta x\ is\ so\ small\ that\ it\ approaches\ 0\\\ \\ \Longrightarrow Slope\ \frac{\Delta y}{\Delta x} =\ \lim_{\Delta x\rightarrow 0}\frac{f( x+\Delta x) -f( x)}{x+\Delta x\ -x}\\\ \\ \Longrightarrow Slope\ \frac{\Delta y}{\Delta x} =\lim_{\Delta x\rightarrow 0} \ \frac{f( x+\Delta x) -f( x)}{\Delta x\ }\\\ \\ This\ can\ also\ be\ written\ as\\\ \\ \Longrightarrow \ \frac{dy}{dx} =\lim _{x\rightarrow 0} \ \frac{f( x+\Delta x)\ -\ f( x)}{\Delta x}\\\ \\

This is the formula for calculate of slope at a particular point.
Remember that slope at a particular point is known as the derivative of the given function. It is basically telling how much y axis is changing with change in value of x (at that particular point).

Derivative Math Problems

(01) Find the derivative at x = 2 of the function f(x) = 3x

Below is the graph of function f(x) = 3x
We have to find the value of slope of function at point A (x=2 & y=6)

The\ slope\ at\ point\ A\ can\ be\ calculated\ as\\\ \\ Slope\ at\ A\ = \lim_{\Delta x\rightarrow 0}\frac{f( x+\Delta x) \ -\ f( x)}{( x+\Delta x) -x}\\\ \\ \ We\ have\ put\ \lim_ {\Delta x\rightarrow 0} \ because\ we\ know\ that\ \Delta x\ is\ so\ small\ that\ it\ approaches\ 0\\\ \\ Slope\ of\ A\ =\ \lim_{\Delta x\rightarrow 0}\frac{3( x+\Delta x) \ -\ 3x}{( \Delta x)}\\\ \\ \ Slope\ of\ A=\lim_{\Delta x\rightarrow 0}\frac{3x+3\Delta x\ -\ 3x}{( \Delta x)}\\\ \\ Slope\ of\ A=\lim _{\Delta x\rightarrow 0}\frac{3\Delta x\ }{( \Delta x)}\\\ \\ Slope\ of\ A\ =3\

Hence the value of slope at Point A (x=2) is 3

(02) Find the derivative of the following function
f(x) = x^2

Solution

we\ know\ that\\\ \\ f(x) \ =x^{2}\\\ \\ let\ us\ find\ the\ value\ of\ function\ at\ x+\Delta x\\\ \\ f(\ x+\Delta x) \ =\ ( \ x+\Delta x)^{2}\\\ \\ Now\ using\ the\ derivative\ formula \\\ \\ \frac{df( x)}{d( x)} =\lim_{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{2} -x^{2}}{\Delta x}\\\ \\ \frac{df( x)}{d( x)} =\lim_{\Delta x\rightarrow 0}\frac{x^{2} +( \Delta x)^{2} +2x.\Delta x-x^{2}}{\Delta x}\\\ \\ \frac{df( x)}{d( x)} =\lim_{\Delta x\rightarrow 0}\frac{( \Delta x)( \Delta x+2x)}{\Delta x}\\\ \\ \frac{df( x)}{d( x)} =\lim_{\Delta x\rightarrow 0}( \Delta x+2x)\\\ \\ using\ the\ limit\ \Delta x=0,\ we\ get\\\ \\ \frac{df( x)}{d( x)} =2x\\\ \\ Hence\ 2x\ is\ the\ derivative\ of\ the\ function\ f( x) =x^{2} \\\ \\

Significance of the Solution
The above solution basically gave us the equation of slope.
With the help of this equation we can find the slope at any point of the given function

For Example
(i) Slope of x^2 at x=2 is:
⟹ putting x=2 in 2x
⟹ 2(2)
⟹ 4
Hence 4 is the slope at point x=2

(ii) Find slope of x^2 at x = 5;
⟹ Putting x=5 in 2x
⟹ 2 (5)
⟹ 10
Hence 10 is the slope at x=5

(03) Find the derivative of following function at x = -1
f( x) \ =2x^{2} +3x\ -\ 5

Solution
we\ know\ that\ \\\ \\ f( x) \ =2x^{2} +3x-5\ \\\ \\ Also\\ \\ f( x+\Delta x) =2( x+\Delta x)^{2} +3( x+\Delta x) -5 \\\ \\ We\ know\ that\\\ \\ \frac{dy}{dx} =\lim_{\Delta x\rightarrow 0}\frac{f( x+\Delta x) -f( x)}{\Delta x}\\\ \\ \frac{dy}{dx} =\lim_ {\Delta x\rightarrow 0}\frac{2( x+\Delta x)^{2} +3( x+\Delta x) -5-2x^{2} -3x+5}{\Delta x}\\\ \\ \frac{dy}{dx} =\lim_{\Delta x\rightarrow 0}\frac{2( x+\Delta x)^{2} +3( x+\Delta x) -2x^{2} -3x}{\Delta x}\\\ \\ \ \frac{dy}{dx} =\lim_{\Delta x\rightarrow 0}\frac{2\left( x^{2} +\Delta x^{2} +2x\Delta x\right) +3x+3\Delta x-2x^{2} -3x}{\Delta x}\\\ \\ \frac{dy}{dx} =\lim_{\Delta x\rightarrow 0}\frac{2x^{2} +2\Delta x^{2} +4x\Delta x+3x+3\Delta x-2x^{2} -3x}{\Delta x}\\\ \\ \ \ \frac{dy}{dx} =\lim_{\Delta x\rightarrow 0}\frac{2\Delta x^{2} +4x\Delta x+3\Delta x}{\Delta x}\\\ \\ \frac{dy}{dx} =\lim _{\Delta x\rightarrow 0} 2\Delta x +4x+3 \\\ \\ putting\ \Delta x\ =0,\\ \\ we\ get\ \frac{dy}{dx} =4x+3\\\ \\

⟹ 4x + 3 is the general equation of the slope
We have to find the slope at x=-1

⟹ 4 (-1) + 3

⟹ – 4 + 3

⟹ -1

Hence -1 is the slope at x = -1 in equation f( x) \ =2x^{2} +3x\ -\ 5