Difference between revisions of "Applications-of-GeoGebra/C3/Differentiation-using-GeoGebra/English"

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−
 
 
{|border=1
 
{|border=1
−
| | '''Visual Cue'''
+
|| '''Visual Cue'''
−
| | '''Narration'''
+
|| '''Narration'''
  
 
|-
 
|-
−
| | '''Slide Number 1'''
+
|| '''Slide Number 1'''
  
 
'''Title Slide'''
 
'''Title Slide'''
−
| | Welcome to this tutorial on '''Differentiation using GeoGebra'''.
+
|| Welcome to this tutorial on '''Differentiation using GeoGebra'''.
 
|-
 
|-
−
| | '''Slide Number 2'''
+
|| '''Slide Number 2'''
  
 
'''Learning Objectives'''
 
'''Learning Objectives'''
−
| | In this tutorial, we will learn how to use '''GeoGebra''' to:
+
|| In this tutorial, we will learn how to use '''GeoGebra''' to:
  
 
Understand Differentiation
 
Understand Differentiation
  
−
Draw graphs of derivative of functions
+
Draw graphs of derivative of functions.
−
 
+
 
|-
 
|-
−
| | '''Slide Number 3'''
+
|| '''Slide Number 3'''
  
 
'''System Requirement'''
 
'''System Requirement'''
−
| | Here I am using:
+
|| Here I am using:
  
−
'''Ubuntu Linux''' OS version 16.04
+
'''Ubuntu Linux''' Operating System version 16.04
  
−
'''GeoGebra''' 5.0.481.0-d
+
'''GeoGebra''' 5.0.481.0-d.
  
 
|-
 
|-
−
| | '''Slide Number 4'''
+
|| '''Slide Number 4'''
  
 
'''Pre-requisites'''
 
'''Pre-requisites'''
−
 
  
 
'''www.spoken-tutorial.org'''
 
'''www.spoken-tutorial.org'''
−
| | To follow this '''tutorial''', you should be familiar with:
+
|| To follow this '''tutorial''', you should be familiar with:
  
 
'''GeoGebra''' interface
 
'''GeoGebra''' interface
Line 45: Line 42:
  
 
|-
 
|-
−
| | '''Slide Number 5'''
+
|| '''Slide Number 5'''
  
 
'''Differentiation: First Principles'''
 
'''Differentiation: First Principles'''
  
−
[[Image:]]
 
  
 
'''f(x) = x<sup>2</sup>-x'''
 
'''f(x) = x<sup>2</sup>-x'''
Line 56: Line 52:
  
 
'''A (x, f(x)), B (x+j, f(x+j))'''
 
'''A (x, f(x)), B (x+j, f(x+j))'''
−
| |
+
||'''Differentiation: First Principles'''
  
 
Let us understand differentiation using '''first principles''' for the '''function f of x'''.
 
Let us understand differentiation using '''first principles''' for the '''function f of x'''.
Line 64: Line 60:
  
 
'''f prime x''' is the derivative of '''f of x'''.
 
'''f prime x''' is the derivative of '''f of x'''.
 +
  
 
Consider 2 points, '''A''' and '''B'''.
 
Consider 2 points, '''A''' and '''B'''.
Line 69: Line 66:
 
'''A''' is '''x''' comma '''f of x''' and '''B''' is '''x''' plus '''j''' comma '''f of x''' plus '''j'''
 
'''A''' is '''x''' comma '''f of x''' and '''B''' is '''x''' plus '''j''' comma '''f of x''' plus '''j'''
 
|-
 
|-
−
| | Show the '''GeoGebra''' window.
+
|| Show the '''GeoGebra''' window.
−
| | I have opened the '''GeoGebra''' interface.
+
|| I have opened the '''GeoGebra''' interface.
 
|-
 
|-
−
| | Type '''f(x)=x^2-x''' in the '''input bar''' >> '''Enter'''
+
|| Type '''f(x)=x^2-x''' in the '''input bar''' >> '''Enter'''
−
| | In the '''input bar''', type the following line and press '''Enter'''.
+
|| In the '''input bar''', type the following line.
  
 
For the '''caret symbol''', hold the '''Shift''' key down and press 6.
 
For the '''caret symbol''', hold the '''Shift''' key down and press 6.
  
−
'''f x''' in parentheses equals '''x caret 2''' minus '''x'''
+
Press '''Enter'''.
 
|-
 
|-
−
| | Point to the equation in '''Algebra''' view.
+
|| Point to the equation in '''Algebra''' view.
  
 
Point to '''parabola''' in '''Graphics''' view.
 
Point to '''parabola''' in '''Graphics''' view.
−
| | The equation appears in the '''Algebra''' view and the '''function f''' is graphed as a '''parabola'''.
+
|| Observe the equation and the parabolic graph of '''function f'''.
 
|-
 
|-
−
| | Point to '''parabola''' in '''Graphics''' view.
+
|| Click on '''Point on Object''' tool >> click on the parabola at '''(2,2)'''.
−
| | It opens upwards and intersects the '''x-axis''' at two points.
+
−
|-
+
−
| | Click on '''Point on Object''' tool >> click on the parabola at '''(2,2)'''.
+
  
 
Point to '''A''' at '''(2,2)'''.
 
Point to '''A''' at '''(2,2)'''.
−
| | Under '''Point''', click on '''Point on Object''' and click on the parabola at 2 comma 2.
 
  
−
This creates point '''A'''.
+
Click on '''Point''' tool and click on '''(3,6)'''.
−
|-
+
 
−
| | Click on '''Point''' tool and click on '''(3,6)'''.
+
|| Clicking on the '''Point on Object''' tool, create point A at 2 comma 2 and B at 3 comma 6.
−
| | Create a point '''B''' at 3 comma 6.
+
 
 
|-
 
|-
−
| | Click on '''Line''' tool and click on points '''B''' and '''A'''.
+
|| Click on '''Line''' tool >> click on points '''B''' and '''A'''.
−
| | Click on '''Line''' tool and click on points '''B''' and '''A'''.
+
|| Click on '''Line''' tool and click on points '''B''' and '''A''' to draw line '''g'''.
 
|-
 
|-
−
| | Click on the '''Move''' tool.
+
|| Click on the '''Move''' tool.
  
 
Double click on the resulting '''line g''' and click on '''Object Properties'''.
 
Double click on the resulting '''line g''' and click on '''Object Properties'''.
Line 107: Line 100:
  
 
Click on '''Style''' tab and select '''dashed style'''.
 
Click on '''Style''' tab and select '''dashed style'''.
−
| | Let us make this line '''g '''blue and dashed.
+
|| As shown earlier in this series, make this line '''g ''' blue and dashed.
 
|-
 
|-
−
| | Click on '''Tangents''' tool under '''Perpendicular Line''' tool.
+
|| Click on '''Tangents''' tool under '''Perpendicular Line''' tool.
−
| | Under '''Perpendicular Line''', click on '''Tangents'''.
+
|| Under '''Perpendicular Line''', click on '''Tangents'''.
 
|-
 
|-
−
| | Click on '''A''' and then on the '''parabola'''.
+
|| Click on '''A''' >> click on the parabola.
−
| | Then click on '''A''' and then on the '''parabola'''.
+
|| Click on '''A''' and then on the parabola.
 
|-
 
|-
−
| | Point to '''tangent h''' at point '''A''' to the '''parabola'''.
+
|| Point to '''tangent h''' at point '''A''' to the parabola.
−
| | This draws a '''tangent h''' at point '''A''' to the '''parabola'''.
+
|| This draws a '''tangent h''' at point '''A''' to the parabola.
 
|-
 
|-
−
| | Double click on '''tangent h''' and click on '''Object Properties'''.
+
|| Double click on '''tangent h''' and click on '''Object Properties'''.
  
 
Under '''Color''' tab, select red.
 
Under '''Color''' tab, select red.
  
 
Close the '''Preferences''' box.
 
Close the '''Preferences''' box.
−
| | Let us make '''tangent h''' a red line.
+
|| Let us make '''tangent h''' a red line.
 
|-
 
|-
−
| | Click on '''Point''' tool and click in '''Graphics''' view.
+
|| Click on '''Point''' tool and click in '''Graphics''' view.
−
| | Click on the '''Point''' tool and click anywhere in '''Graphics''' view.
+
−
|-
+
−
| | Point to point '''C'''.
+
−
| | This creates point '''C'''.
+
−
|-
+
−
| | Double click on point '''C''' in '''Algebra''' view and change its '''coordinates''' to '''(x(B),y(A))'''.
+
−
| | In '''Algebra''' view, double-click on '''C''' and change its '''coordinates''' to the following ones.
+
  
−
In parentheses, '''x B''' in parentheses comma '''y A''' in parentheses.
+
Point to point '''C'''.
 +
|| Click on the '''Point''' tool and click anywhere in '''Graphics''' view.
  
−
Press '''Enter'''.
+
This creates point '''C'''.
 
|-
 
|-
−
| | Point to '''C'''.
+
|| Double click on point '''C''' in '''Algebra''' view and change its '''coordinates''' to '''(x(B),y(A))'''.
−
| | Now C has the same '''x coordinate''' as point '''B''' and the same '''y coordinate''' as point '''A'''.
+
 
 +
Point to '''C'''.
 +
|| In '''Algebra''' view, double-click on '''C''' and change its '''coordinates''' to the following.
 
|-
 
|-
−
| |
+
||Point to B and A.
−
| | Let us draw segments '''BC''' and '''AC.'''
+
|| Now C has the same '''x coordinate''' as point '''B''' and the same '''y coordinate''' as point '''A'''.
 
|-
 
|-
−
| | Under '''Line''', click on '''Segment''' and click on '''B '''and '''C''', and then on '''A''' and '''C'''.
+
|| Under '''Line''', click on '''Segment''' and click on '''B '''and '''C''', and then on '''A''' and '''C'''.
−
| | Under '''Line''', click on '''Segment''' and click on '''B''' and '''C'''.
+
|| Let us use the '''Segment''' tool to draw segments '''BC''' and '''AC'''.
 +
|-
 +
|| Right-click on '''C''' >> Select '''Object Properties''' >> '''Color''' tab >> Purple
  
−
Then, click on '''A''' and '''C''' to draw '''AC'''.
+
Click on '''Style''' tab >> select dashed line
  
−
Note that '''BC''' and '''AC''' are called '''i''' and '''j''' in the order of their creation.
+
Under '''Basic''' tab >> choose '''Name and Value''' >> '''Show Label''' check box.
−
|-
+
−
| | Right-click on '''AC''' >> Select '''Object Properties''' >> '''Color''' tab >> Purple
+
  
−
Click on '''Style''' tab >> select dashed line
+
Close the '''Preferences''' dialog box.
−
| | We will make '''AC''' and '''BC''' purple and dashed segments.
+
|| We will make '''AC''' and '''BC''' purple and dashed segments.
 
|-
 
|-
−
| | Under '''Basic''' tab >> choose '''Name and Value''' >> '''Show Label''' check box.
+
|| With '''Move''' highlighted, drag '''B''' towards '''A''' on the parabola.
−
| | Under '''Basic''' tab, choose '''Name and Value''' from the dropdown menu next to the '''Show Label''' check box.
+
|| With '''Move''' highlighted, drag '''B''' towards '''A''' on the parabola.
 
|-
 
|-
−
| | Close the '''Preferences''' dialog box.
+
|| Point to the value of '''j''' (length of '''AC''') and lines '''g''' and '''h'''.
−
| | Close the '''Preferences''' dialog box.
+
|| Observe lines '''g''' and '''h''' and the value of '''j''' (length of '''AC''').
−
|-
+
−
| | Click on '''Move''' and drag '''B''' towards '''A''' on the '''parabola'''.
+
−
| | Click on '''Move''' and drag '''B''' towards '''A''' on the '''parabola'''.
+
−
|-
+
−
| | Point to the value of '''j''' (length of '''AC''') and lines '''g''' and '''h'''.
+
−
| | Observe lines '''g''' and '''h''' and the value of '''j''' (length of '''AC''').
+
  
−
As '''j''' approaches 0, points '''B'''and '''A''' begin to overlap.
+
As '''j''' approaches 0, points '''B''' and '''A''' begin to overlap.
  
 
Lines '''g''' and '''h''' also begin to overlap.
 
Lines '''g''' and '''h''' also begin to overlap.
 
|-
 
|-
−
| | Point to line '''g''', '''BC''' and '''AC'''.
+
|| Point to line '''g''', '''BC''' and '''AC'''.
−
| | Slope of line '''g''' is the ratio of length of '''BC''' to length of '''AC'''.
+
|| Slope of line '''g''' is the ratio of length of '''BC''' to length of '''AC'''.
 
|-
 
|-
−
| | Point to all the points on the parabola.
+
|| Point to all the points on the parabola.
−
| | Derivative of the parabola is the slopes of tangents at all points on curve.
+
|| Derivative of the parabola is the slope of the tangent at each point on the curve.
 
|-
 
|-
−
| | Point to the lines '''g''' and '''h'''.
+
|| Point to text-box that appears in '''GeoGebra''' window.
−
| | When '''j''' equals 0, line '''g''' through '''A''' and '''B''' coincides with the tangent line '''h''' at '''A'''.
+
−
 
+
−
But slope of '''g''' is undefined as the denominator of the ratio of '''i''' to '''j''' is 0.
+
−
|-
+
−
| | Point to text-box that appears in '''GeoGebra''' window.
+
  
 
As '''B''' approaches '''A''', slope '''AB''' approaches slope of tangent at '''A'''.
 
As '''B''' approaches '''A''', slope '''AB''' approaches slope of tangent at '''A'''.
−
| | As '''B''' approaches '''A''' on '''f of x''', slope of '''AB''' approaches the slope of tangent at '''A'''.
+
|| As '''B''' approaches '''A''' on '''f of x''', slope of '''AB''' approaches the slope of tangent at '''A'''.
 
|-
 
|-
−
| |
+
||
−
| | Now let us look at the '''Algebra''' behind these concepts.
+
|| Now let us look at the '''Algebra''' behind these concepts.
 
|-
 
|-
−
| | '''Slide Number 7'''
+
|| '''Slide Number 6'''
  
 
'''Differentiation: First Principles, the Algebra'''
 
'''Differentiation: First Principles, the Algebra'''
  
−
'''f'(x) = lim_(j→0) length of Segment BC/length of Segment AC'''
 
  
−
.......''' = lim_(j→0) [(f(x+j) – f(x)]/[(x+j) – x]'''
+
'''f'(x) = lim_j→0  (length of Segment BC / length of Segment AC)'''
 +
 
 +
'''<nowiki>= lim_j→0 </nowiki>[(f(x+j) – f(x)]/[(x+j) – x]'''
  
 
Remember '''f(x) = x<sup>2</sup>-x, (x+j)<sup>2</sup> = x<sup>2</sup>+2xj+j<sup>2</sup>'''
 
Remember '''f(x) = x<sup>2</sup>-x, (x+j)<sup>2</sup> = x<sup>2</sup>+2xj+j<sup>2</sup>'''
  
−
'''f'(x) = lim_(j→0) [(x+j)<sup>2</sup>-(x+j)-(x<sup>2</sup>-x)]/[(x+j-x]'''
+
'''f'(x) = lim_j→0 [(x+j)<sup>2</sup>-(x+j)-(x<sup>2</sup>-x)]/(x+j-x)'''
 +
         
 +
||'''Differentiation: First Principles, the Algebra'''
  
−
| | Slope of line '''AB''' equals the ratio of the lengths of '''BC''' to '''AC'''.
+
Slope of line '''AB''' equals the ratio of the lengths of '''BC''' to '''AC'''.
  
 
Line '''AB''' becomes the tangent at point '''A''' as distance '''j''' between '''A''' and '''B''' approaches 0.
 
Line '''AB''' becomes the tangent at point '''A''' as distance '''j''' between '''A''' and '''B''' approaches 0.
  
−
'''BC''' is the difference between '''y' coordinates''', '''f of x''' plus '''j''' and '''f of x''', for '''A''' and '''B'''.
+
 
 +
'''BC''' is the difference between '''y coordinates''', '''f of x''' plus '''j''' and '''f of x''', for '''A''' and '''B'''.
 +
 
  
 
'''AC''' is the difference between the '''x-coordinates''', '''x''' plus '''j''' and '''x'''.
 
'''AC''' is the difference between the '''x-coordinates''', '''x''' plus '''j''' and '''x'''.
Line 215: Line 198:
 
We will expand the terms in the numerator.
 
We will expand the terms in the numerator.
 
|-
 
|-
−
| | '''Slide 8 Differentiation: First Principles—the Algebra-Cont’d'''
+
|| '''Slide Number 7
  
−
'''f'(x) = lim_(j→0) [x<sup>2</sup>+2xj+j<sup>2</sup>-x-j-x<sup>2</sup>+x]/j'''
+
'''The Algebra-Cont’d'''
  
−
..............'''= lim_(j→0) [2xj+j<sup>2</sup>-j]/j = lim_(j→0) [j(2x+j-1)]/j'''
 
  
−
..................'''= lim_(j→0) [2x+j-1] = 2x-1'''
+
'''f'(x) = lim_j→0 [x<sup>2</sup>+2xj+j<sup>2</sup>-x-j-x<sup>2</sup>+x]/j'''
  
−
'''f'(x<sup>2</sup>-x) = 2x - 1'''
+
'''<nowiki>= lim_j→0 [</nowiki>2xj+j<sup>2</sup>-j]/j = lim_j→0 [j(2x+j-1)]/j'''
−
| | After expanding the terms in the numerator, we will cancel out similar terms with opposite signs.
+
  
−
As '''j''' is a common factor in the numerator, we will pull it out.
 
  
−
Now, we can cancel '''j''' from both the numerator and denominator.
+
'''<nowiki>= lim_j→0 [2x+j-1] = 2x-1</nowiki>'''
 +
 
 +
'''f'(x<sup>2</sup>-x) = 2x -1'''
 +
|| After expanding the terms in the numerator, we will cancel out similar terms with opposite signs.
 +
 
 +
 
 +
We will pull out '''j''' from the numerator and cancel it.
 +
 
 +
 
 +
Note that as '''j''' approaches 0, '''j''' can be ignored. So that '''2x''' plus '''j''' minus 1 approaches '''2x''' minus 1.
  
−
Note that as '''j''' approaches 0, '''j''' can be ignored so that '''2x''' plus '''j''' minus  1 approaches '''2x''' minus 1.
 
  
 
As we know, derivative of '''x squared''' minus<sup> '''</sup>x''' is '''2x''' minus 1.
 
As we know, derivative of '''x squared''' minus<sup> '''</sup>x''' is '''2x''' minus 1.
  
−
Thus, the derivative of a '''function''' is the slope of the tangent at a point on the function.
 
 
|-
 
|-
−
| |
+
||
−
| | Let us look at derivative graphs for some '''functions'''.
+
|| Let us look at derivative graphs for some '''functions'''.
 
|-
 
|-
−
| | '''Slide Number 10'''
+
|| '''Slide Number 8'''
  
 
'''Differentiation of a Polynomial Function'''
 
'''Differentiation of a Polynomial Function'''
 +
  
 
Consider '''g(x)=5+12x-x<sup>3'''</sup>
 
Consider '''g(x)=5+12x-x<sup>3'''</sup>
  
−
'''d(5+12x-x<sup>3</sup>)/dx = d(5)/dx + d(12x)/dx - d(x<sup>3</sup>)/dx = 0 + 12 - 3x <sup>2 </sup> = -3x <sup>2 </sup>+12'''
+
'''Differentiation rules''':
 +
 
 +
'''<div>d(u±v)/dx = d(u)/dx ± d(v)/dx</div>'''
 +
 
 +
'''d(5+12x-x<sup>3</sup>)/dx = d(5)/dx + d(12x)/dx - d(x<sup>3</sup>)/dx <nowiki>= 0 + 12 - 3x</nowiki><sup>2 </sup><nowiki>= -3x</nowiki><sup>2 </sup>+12'''
 +
 
  
 
For '''g(x)=5+12x-x<sup>3</sup>, g'(x) = -3x<sup>2 </sup>+12'''
 
For '''g(x)=5+12x-x<sup>3</sup>, g'(x) = -3x<sup>2 </sup>+12'''
−
| |
+
||'''Differentiation of a Polynomial Function'''
−
Consider '''g of x''' equals 5 plus '''12 x '''minus '''x cubed'''.
+
 
 +
Consider '''g of x'''.
  
 
Derivative '''g prime x''' is the sum and difference of derivatives of the individual components.
 
Derivative '''g prime x''' is the sum and difference of derivatives of the individual components.
  
−
'''g prime x''' equals 5 plus '''12 x''' minus '''x cubed''' is calculated by applying these rules.
+
 
 +
'''g prime x''' is calculated by applying these rules.
 
|-
 
|-
−
| |
+
||
−
| | Let us differentiate '''g of x''' in '''GeoGebra'''.
+
|| Let us differentiate '''g of x''' in '''GeoGebra'''.
 
|-
 
|-
−
| | Open a new '''GeoGebra''' window.
+
|| Open a new '''GeoGebra''' window.
−
| | Open a new '''GeoGebra''' window.
+
|| Open a new '''GeoGebra''' window.
 
|-
 
|-
−
| | Type '''g(x)=5+12x-x^3''' in '''input bar''' >> '''Enter'''
+
|| Type '''g(x)=5+12x-x^3''' in '''input bar''' >> '''Enter'''
−
| | In the '''input bar''', type the following line and press '''Enter'''.
+
|| In the '''input bar''', type the following line and press '''Enter'''.
 +
|-
 +
|| Under '''Move Graphics View''', click on '''Zoom Out'''.
 +
 
 +
Click in '''Graphics''' view until you see '''function g'''.
 +
 
 +
|| As shown earlier in the series, zoom out to see '''function g''' properly.  
  
−
'''g x''' in parentheses equals 5 plus '''12 x''' minus '''x caret''' 3
 
 
|-
 
|-
−
| | Right-click in '''Graphics''' view and select '''xAxis : yAxis''' option.
+
|| Right-click in '''Graphics''' view and select '''xAxis : yAxis''' option.
  
 
Select '''1:5'''.
 
Select '''1:5'''.
−
| | Right-click in '''Graphics''' view and select '''xAxis''' is to '''yAxis''' option.
+
|| Right-click in '''Graphics''' view and select '''xAxis''' is to '''yAxis''' option.
  
 
Select 1 is to 5.
 
Select 1 is to 5.
 
|-
 
|-
−
| | Click on '''Point on Object''' tool and click on the curve to create point '''A'''.
+
|| Under '''Move Graphics View''', click on '''Zoom Out''' again.  
−
| | Click on '''Point on Object''' tool and click on curve '''g of x''' to create point '''A'''.
+
 
 +
Click in '''Graphics''' view to zoom out. 
 +
|| I will zoom out again.  
 
|-
 
|-
−
| | Click on '''Tangent''' under '''Perpendicular Line'''.
+
|| Click on '''Point on Object''' tool and click on the curve to create point '''A'''.
 +
 
 +
Click on '''Tangent''' under '''Perpendicular Line'''.
  
 
Click on point '''A''' and the curve.
 
Click on point '''A''' and the curve.
−
| | Under '''Perpendicular Line''', click on '''Tangent'''.
 
  
−
Now click on point '''A''' and the curve '''g of x'''.
+
|| As shown earlier, draw point '''A''' on curve '''g''' and a tangent '''f''' at this point.  
 
|-
 
|-
−
| | Point to tangent line '''f''' to the curve at point '''A'''.
+
|| Click on '''Slope''' tool under '''Angle''' tool and on tangent line '''f'''.
−
| | This draws a tangent line '''f''' to the curve at point '''A'''.
+
|| Under '''Angle''', click on '''Slope''' and on tangent line '''f'''.
 
|-
 
|-
−
| | Click on '''Slope''' tool under '''Angle''' tool and on tangent line '''f'''.
+
|| Point to slope of '''line f''' at '''A''' appearing as '''m''' value in '''Graphics''' view.
−
| | Under '''Angle''', click on '''Slope''' and on tangent line '''f'''.
+
|| Slope of tangent line '''f''' appears as '''m''' value in '''Graphics''' view.
 
|-
 
|-
−
| | Point to slope of '''line f''' at '''A''' appearing as '''m''' value in '''Graphics''' view.
+
|| Click on '''Point''' tool and in '''Graphics''' view to create point '''B'''.
−
| | Slope of line '''f'''at '''A''' appears as '''m''' value in '''Graphics''' view.
+
−
|-
+
−
| | Click on '''Point''' tool and in '''Graphics''' view to create point '''B'''.
+
−
| | Click on '''Point''' tool and click in '''Graphics''' view to create point '''B'''.
+
−
|-
+
−
| | Double click on point '''B''' in '''Algebra''' view and change '''coordinates''' to ('''x(A), m)'''.
+
−
| | In '''Algebra''' view, double click on '''B''' and change its '''coordinates''' to '''x A''' in parentheses comma '''m'''.
+
  
−
Press '''Enter'''.
+
Double click on point '''B''' in '''Algebra''' view and change '''coordinates''' to ('''x(A), m)'''.
−
|-
+
−
| | Point to points '''A''' and '''B''' and slope ''' m''' of tangent line '''g'''.
+
−
| | This makes '''x coordinate''' of '''B''' the same as that of '''A'''.
+
  
−
And slope '''m''' of tangent line '''f''' is its '''y coordinate'''.
+
Point to points '''A''' and '''B''' and slope ''' m''' of tangent line '''g'''.
 +
 
 +
|| Draw point '''B''' and change its '''coordinates''' to '''x A''' in parentheses comma '''m'''.
 
|-
 
|-
−
| | Right-click on point '''B''' and select '''Trace On''' option.
+
|| Right-click on point '''B''' >> select '''Trace On''' option.
−
| | Right-click on point '''B''' and select '''Trace On''' option
+
|| Right-click on '''B''' and select '''Trace On''' option
 
|-
 
|-
−
| | Click on '''Move''' tool and move point '''A''' on curve.
+
|| Click on '''Move''' tool and move point '''A''' on curve.
  
 
Observe the curve traced by point '''B'''.
 
Observe the curve traced by point '''B'''.
−
| | Click on '''Move''' tool and move point '''A''' on the curve.
+
|| With '''Move''' tool highlighted, move point '''A''' on curve.
  
 
Observe the curve traced by point '''B'''.
 
Observe the curve traced by point '''B'''.
 
|-
 
|-
−
| | Type '''Deri''' in '''input bar''' >> select '''Derivative( <Function> )''' >> Type '''g''' instead of highlighted '''<Function>''' >> press '''Enter'''
+
||
−
| | In the '''input bar''', type '''capital D e r i'''.
+
|| Let us check whether we have the correct '''derivative''' graph.
 +
|-
 +
|| Type '''Deri''' in '''input bar''' >> select '''Derivative( <Function> )''' >> Type '''g''' instead of highlighted '''<Function>''' >> press '''Enter'''
 +
|| In the '''input bar''', type '''d e r i'''.
 +
 
  
 
From the menu that appears, select '''Derivative Function''' option.
 
From the menu that appears, select '''Derivative Function''' option.
  
 
Type '''g''' to replace the highlighted word '''<Function>'''.
 
Type '''g''' to replace the highlighted word '''<Function>'''.
 +
  
 
Press '''Enter'''.
 
Press '''Enter'''.
 
|-
 
|-
−
| | Point to the derivative graph.
+
||Point to the equation in '''Algebra''' view.
−
| | This will confirm that you have the correct derivative graph.
+
 
 +
Drag the boundary.  
 +
|| Note the equation of '''g prime x''' in '''Algebra''' view.
 +
 
 +
Drag the boundary to see it properly
 
|-
 
|-
−
| | Point to the equation of '''g prime x''' in '''Algebra''' view.
+
|| Compare slides' calculations with equation of '''g'(x)''' in '''Algebra''' view.
−
| | Note the equation of '''g prime x''' in '''Algebra''' view.
+
|| Compare the calculations in the previous slide with the equation of '''g prime x'''
 
|-
 
|-
−
| | Drag the boundary to see it properly.
+
||
−
| | Drag the boundary to see it properly.
+
|| Let us find the maxima and minima of the '''function g of x'''.
−
|-
+
−
| | Compare slide’s calculations with equation of '''g'(x)''' in '''Algebra''' view.
+
−
| | Compare the calculations in the previous slide with the equation of '''g prime x'''
+
−
|-
+
−
| |
+
−
| | Let us use differentiation to find the maxima and minima of a '''function'''.
+
−
 
+
−
Let us look at '''g of x'''.
+
 
|-
 
|-
−
| | Point to derivative curve '''g'(x)''' above the '''x-axis''' and to '''g(x)'''.
+
|| Point to derivative curve '''g'(x)''' above the '''x-axis''' and to '''g(x)'''.
−
| | Derivative curve '''g prime x''' remains above the '''x-axis''' (is positive) as long as '''g of x''' is increasing.
+
|| Derivative curve '''g prime x''' remains above the '''x-axis''' (is positive) as long as '''g of x''' is increasing.
 
|-
 
|-
−
| | Point to derivative curve '''g'(x)''' below the '''x-axis''' and to '''g(x)'''.
+
|| Point to derivative curve '''g'(x)''' below the '''x-axis''' and to '''g(x)'''.
−
| | '''g prime x''' remains below the '''x-axis''' (is negative) as long as '''g of x''' is decreasing.
+
|| '''g prime x''' remains below the '''x-axis''' (is negative) as long as '''g of x''' is decreasing.
 
|-
 
|-
−
| | Point to derivative curve '''g'(x)''' intersecting '''x-axis''' at '''x = -2 '''and''' x = 2'''.
+
|| Point to derivative curve '''g'(x)''' intersecting '''x-axis''' at '''x = -2 '''and''' x = 2'''.
−
| | 2 and -2 are the values of '''x''' when '''g prime x''' equals 0.
+
|| 2 and -2 are the values of '''x''' when '''g prime x''' equals 0.
 
|-
 
|-
−
| |
+
||Point to slope.
−
| | Slope of the tangents at the corresponding points on '''g of x''' is 0.
+
|| Slope of the tangent at the corresponding point on '''g of x''' is 0.
  
−
These points on '''g of x''' are maxima or a minima.
+
Such points on '''g of x''' are maxima or minima.
 
|-
 
|-
−
| |
+
||Point to '''(-2,-11)''' and '''(2,21)'''.
−
 
+
|| Hence, for '''g of x,''' -2 comma -11 is the minimum and 2 comma 21 is the maximum.
−
Point to '''(-2,-11)''' and '''(2,21)'''.
+
−
| | Hence, for '''g of x,''' -2 comma -11 is the minimum and 2 comma 21 is the maximum.
+
 
|-
 
|-
−
| | Point to minimum of '''g(x)''' and '''x=-3''' and '''x = -1'''.
+
|| Point to minimum of '''g(x)''' and '''x=-3''' and '''x = -1'''.
−
| | In '''GeoGebra''', we can see that the minimum value of '''g of x''' lies between '''x''' equals -3 and '''x''' equals -1.
+
|| In '''GeoGebra''', we can see that the minimum value of '''g of x''' lies between '''x''' equals -3 and '''x''' equals -1.
 
|-
 
|-
−
| | In the '''input bar''', type '''Min''' with capital M.
+
|| In the '''input bar''', type '''Min'''.
  
−
From the menu that appears, select '''Min Function Start x-Value End x-Value''' option.
+
From the menu that appears, select '''Min Function Start x-Value, End x-Value''' option.
  
−
Type '''g''' instead of the highlighted word '''Function'''.
+
Type '''g''' for '''Function'''.
  
−
Press '''Tab''' to move to and highlight '''Start x-Value''' and type -3.
+
Press '''Tab''' to go to the next argument.
  
−
Again, press '''Tab''' to move to and highlight '''End x-Value''' and type -1.
+
Type -4 and -1 as '''Start''' and '''End x-Values'''.
  
 
Press '''Enter'''.
 
Press '''Enter'''.
−
| | In the '''input bar''', type '''Min''' with capital M.
+
|| In the '''input bar''', type '''Min'''.
  
−
From the menu that appears, select '''Min Function Start x-Value End x-Value''' option.
+
From the menu that appears, select '''Min Function Start x-Value, End x-Value''' option.
  
−
Type '''g''' instead of the highlighted word '''Function'''.
+
Type '''g''' for '''Function'''.
  
−
Press '''Tab''' to move to and highlight '''Start x-Value''' and type -3.
+
Press '''Tab''' to go to the next argument.
  
−
Again, press '''Tab''' to move to and highlight '''End x-Value''' and type -1.
+
Type -4 and -1 as '''Start''' and '''End x-Values'''.
  
 
Press '''Enter'''.
 
Press '''Enter'''.
 
|-
 
|-
−
| | Point to minimum '''C''' in '''Graphics '''view and its '''co-ordinates''' in '''Algebra''' view.
+
|| Point to minimum '''C''' in '''Graphics''' view and its '''co-ordinates''' in '''Algebra''' view.
−
| | In '''Graphics''' view, we see the minimum on '''g of x'''.
+
|| In '''Graphics view''', we see the minimum on '''g of x'''.
  
−
Its '''co-ordinates''' are -2 comma -11 in '''Algebra''' view.
+
Its '''co-ordinates''' are -2 comma -11 in '''Algebra''' '''view'''.
 
|-
 
|-
−
| | In the '''input bar''', type '''Max''' with capital M.
+
|| In the '''input bar''', type '''Max'''.
  
−
From the menu that appears, select '''Max Function Start x-Value End x-Value''' option.
+
From the menu that appears, select '''Max Function Start x-Value, End x-Value''' option.
  
−
Type '''g''' instead of the highlighted word '''Function'''.
+
Type '''g''', 1 and 4 as the arguments.
−
 
+
−
Press '''Tab''' to move to and highlight '''Start x-Value''' and type 1.
+
−
 
+
−
Again, press '''Tab''' to move to and highlight '''End x-Value''' and type 4.
+
  
 
Press '''Enter.'''
 
Press '''Enter.'''
−
| | In the '''input bar''', type '''Max''' with capital M.
+
|| In the '''input bar''', type '''Max'''.
  
−
From the menu that appears, select '''Max Function Start x-Value End x-Value''' option.
+
From the menu that appears, select '''Max Function Start x-Value, End x-Value''' option.
  
−
Type '''g''' instead of the highlighted word '''Function'''.
 
  
−
Press '''Tab''' to move to and highlight '''Start x-Value''' and type 1.
+
Type '''g''', 1 and 4 as the arguments.
  
−
Again, press '''Tab''' to move to and highlight '''End x-Value''' and type 4.
+
Press '''Enter.'''
−
 
+
−
Press '''Enter'''.
+
 
|-
 
|-
−
| | Point to maximum '''C''' in '''Graphics''' view and its '''co-ordinates''' in '''Algebra''' view.
+
|| Point to maximum '''C''' in '''Graphics''' view and its '''co-ordinates''' in '''Algebra''' view.
−
| | In '''Graphics''' view, we see the maximum on '''g of x'''.
+
|| We see the maximum on '''g of x''', 2 comma 21.
−
 
+
−
Its '''co-ordinates''' are 2 comma 21 in '''Algebra''' view.
+
 
|-
 
|-
−
| |
+
||
−
| | Finally, let us take a look at a practical application of differentiation.
+
|| Finally, let us take a look at a practical application of differentiation.
 
|-
 
|-
−
| | '''Slide Number 16'''
+
|| '''Slide Number 9'''
  
 
'''A Practical Application of Differentiation'''
 
'''A Practical Application of Differentiation'''
 +
  
 
We have a 24 inches by 15 inches piece of cardboard
 
We have a 24 inches by 15 inches piece of cardboard
Line 430: Line 416:
 
We have to convert it into a box
 
We have to convert it into a box
  
−
Squares have to be cut from the four corners
+
Squares have to be cut out from the four corners
  
 
What size squares should we cut out to get the maximum volume of the box?
 
What size squares should we cut out to get the maximum volume of the box?
−
| | '''A Practical Application of Differentiation'''
+
|| We have a 24 inches by 15 inches piece of cardboard.
−
 
+
−
We have a 24 inches by 15 inches piece of cardboard.
+
  
 
We have to convert it into a box.
 
We have to convert it into a box.
Line 443: Line 427:
 
What size squares should we cut out to get the maximum volume of the box?
 
What size squares should we cut out to get the maximum volume of the box?
 
|-
 
|-
−
| | '''Slide Number 17'''
+
|| '''Slide Number 10'''
  
 
'''A Sketch of the Cardboard'''
 
'''A Sketch of the Cardboard'''
  
 
Let’s draw the cardboard:
 
Let’s draw the cardboard:
−
 
−
[[Image:]]
 
  
 
The volume function here is '''(24-2x)*(15-2x)*x''' cubic inches.
 
The volume function here is '''(24-2x)*(15-2x)*x''' cubic inches.
−
| | '''A Sketch of the Cardboard'''
+
|| '''A Sketch of the Cardboard'''
  
 
Let us draw the cardboard:
 
Let us draw the cardboard:
Line 460: Line 442:
 
You could expand it into a '''cubic polynomial'''<nowiki>; but we will leave it as it is. </nowiki>
 
You could expand it into a '''cubic polynomial'''<nowiki>; but we will leave it as it is. </nowiki>
 
|-
 
|-
−
| | Open a new '''GeoGebra''' window.
+
|| Open a new '''GeoGebra''' window.
−
| | Open a new '''GeoGebra''' window.
+
|| Open a new '''GeoGebra''' window.
 
|-
 
|-
−
| | Type '''(24-2 x) (15-2 x) x''' in the '''input bar''' >> '''Enter'''.
+
|| Type '''(24-2 x) (15-2 x) x''' in the '''input bar''' >> '''Enter'''.
−
| | In the '''input bar''', type the following line and press '''Enter'''.
+
|| In the '''input bar''', type the following line and press '''Enter'''.
−
 
+
−
In parentheses, 24 minus 2 space '''x''' space in parentheses 15 minus 2 space '''x''' space '''x'''
+
 
|-
 
|-
−
| | Drag the boundary to see the equation properly in '''Algebra''' view.
+
|| Drag the boundary to see the equation properly in '''Algebra''' view.
−
| | Drag the boundary to see the equation properly in '''Algebra''' view.
+
|| Drag the boundary to see the equation properly in '''Algebra''' view.
 
|-
 
|-
−
| | Right-click in '''Graphics''' view and set '''xAxis : yAxis''' to '''1:50'''.
+
|| Right-click in '''Graphics''' view and set '''xAxis : yAxis''' to '''1:50'''.
−
| | Right-click in '''Graphics''' view and set '''xAxis''' is to '''yAxis''' to 1 is to 50.
+
 
 +
Under '''Move Graphics View''', click on '''Zoom Out'''.
 +
 
 +
Click in '''Graphics''' view to see the '''function''' properly. 
 +
 
 +
|| Right-click in '''Graphics''' view and set '''xAxis''' is to '''yAxis''' to 1 is to 50.
 +
 
 +
Now, zoom out to see the function properly. 
 
|-
 
|-
−
| | Point to the graph for this volume '''function''' in '''Graphics''' view.
+
|| Point to the graph for this volume '''function''' in '''Graphics''' view.
  
−
Click in and drag the background to move '''Graphics''' view to see the maximum.
 
−
| | Observe the graph that is plotted for this volume '''function''' in '''Graphics''' view.
 
  
 
Click in and drag the background to move '''Graphics''' view to see the maximum.
 
Click in and drag the background to move '''Graphics''' view to see the maximum.
 +
|| Observe the graph that is plotted for the volume '''function''' in '''Graphics''' view.
 +
 +
Drag the background to see the maximum.
 
|-
 
|-
−
| | Point to the maximum on top of the broad peak and to '''x''' = 0 and '''x''' = 7.
+
|| Point to the maximum on top of the broad peak and to '''x''' = 0 and '''x''' = 7.
−
| | Note that the maximum is on the top of a broad peak from '''x''' equals 0 to '''x''' equals 7.
+
|| Note that the maximum is on the top of this broad peak.
 
|-
 
|-
−
| | Point to both axes.
+
|| Point to both axes.
−
| | The length of the square side is plotted along the '''x-axis'''.
+
|| The length of the square side is plotted along the '''x-axis'''.
  
 
Volume of the box is plotted along the '''y-axis'''.
 
Volume of the box is plotted along the '''y-axis'''.
 
|-
 
|-
−
| | In the '''input bar''', type '''Max''' with capital M.
+
|| In the '''input bar''', type '''Max''' with capital M.
 +
 
  
 
From the menu that appears, select '''Max Function Start x-Value End x-value'''.
 
From the menu that appears, select '''Max Function Start x-Value End x-value'''.
 +
  
 
Instead of highlighted '''Function''', type '''f'''.
 
Instead of highlighted '''Function''', type '''f'''.
 +
  
 
Press '''Tab''' to move and highlight '''Start x-Value''' and type 0.
 
Press '''Tab''' to move and highlight '''Start x-Value''' and type 0.
 +
  
 
Again, press '''Tab''' to move and highlight '''End x-Value''' and type 10.
 
Again, press '''Tab''' to move and highlight '''End x-Value''' and type 10.
 +
  
 
Press '''Enter'''.
 
Press '''Enter'''.
−
| | As before, let us find the maximum of this '''function'''.
+
|| As before, let us find the maximum of this '''function'''.
 +
 
 
|-
 
|-
−
| | Point to the maximum,''' A''', in '''Graphics''' view and its '''coordinates''' in '''Algebra''' view.
+
||Point to the maximum,''' A''', in '''Graphics''' view and its '''coordinates''' in '''Algebra''' view.
−
| | This maps the maximum, point '''A''', on the curve.
+
|| This maps the maximum, point '''A''', on the curve.
−
 
+
−
Click in and drag the background to see
+
  
 
Its '''coordinates''' 3 comma 486 appear in '''Algebra''' view.
 
Its '''coordinates''' 3 comma 486 appear in '''Algebra''' view.
  
−
Thus, we have to cut out 3 inches squares from all corners.
+
 
 +
Thus, we have to cut out 3 inch squares from all corners.
  
 
This will give the maximum possible volume of 486 cubic inches for the cardboard box.
 
This will give the maximum possible volume of 486 cubic inches for the cardboard box.
 
|-
 
|-
−
| |
+
||
−
| | Let us summarize.
+
|| Let us summarize.
 
|-
 
|-
−
| | '''Slide Number 19'''
+
|| '''Slide Number 11'''
  
 
'''Summary'''
 
'''Summary'''
−
| | In this '''tutorial''', we have learnt how to use '''GeoGebra''' to:
+
|| In this tutorial, we have learnt how to use '''GeoGebra''' to:
  
 
Understand differentiation
 
Understand differentiation
Line 525: Line 518:
 
Draw graphs of derivatives of '''functions'''
 
Draw graphs of derivatives of '''functions'''
 
|-
 
|-
−
| | '''Slide Number 16'''
+
|| '''Slide Number 12'''
  
 
'''Assignment'''
 
'''Assignment'''
Line 538: Line 531:
  
 
Find the derivatives of these '''functions''' independently and compare with '''GeoGebra''' graphs.
 
Find the derivatives of these '''functions''' independently and compare with '''GeoGebra''' graphs.
−
| | As an assignment:
+
|| As an assignment:
  
 
Draw graphs of derivatives of the following functions in '''GeoGebra'''.
 
Draw graphs of derivatives of the following functions in '''GeoGebra'''.
 +
  
 
Find the derivatives of these '''functions''' independently and compare with '''GeoGebra''' graphs.
 
Find the derivatives of these '''functions''' independently and compare with '''GeoGebra''' graphs.
 
|-
 
|-
−
| | '''Slide Number 17'''
+
|| '''Slide Number 13'''
  
 
'''About Spoken Tutorial project'''
 
'''About Spoken Tutorial project'''
−
| | The video at the following link summarizes the '''Spoken Tutorial''' project.
+
|| The video at the following link summarizes the '''Spoken Tutorial''' project.
  
 
Please download and watch it.
 
Please download and watch it.
 
|-
 
|-
−
| | '''Slide Number 18'''
+
|| '''Slide Number 14'''
  
 
'''Spoken Tutorial workshops'''
 
'''Spoken Tutorial workshops'''
−
| | The '''Spoken Tutorial Project '''team:
+
|| The '''Spoken Tutorial Project '''team:
  
 
Conducts workshops using spoken tutorials and
 
Conducts workshops using spoken tutorials and
Line 561: Line 555:
 
For more details, please write to us.
 
For more details, please write to us.
 
|-
 
|-
−
| | '''Slide Number 19'''
+
|| '''Slide Number 15'''
  
 
'''Forum for specific questions:'''
 
'''Forum for specific questions:'''
Line 574: Line 568:
  
 
Someone from our team will answer them
 
Someone from our team will answer them
−
| | Please post your timed queries on this forum.
+
|| Please post your timed queries on this forum.
 
|-
 
|-
−
| | '''Slide Number 20'''
+
|| '''Slide Number 16'''
  
 
'''Acknowledgement'''
 
'''Acknowledgement'''
−
| | '''Spoken Tutorial Project''' is funded by NMEICT, MHRD, Government of India.
+
|| '''Spoken Tutorial Project''' is funded by NMEICT, MHRD, Government of India.
  
 
More information on this mission is available at this link.
 
More information on this mission is available at this link.
 
|-
 
|-
−
| |
+
||
−
| | This is '''Vidhya Iyer''' from''' IIT Bombay''', signing off.
+
|| This is '''Vidhya Iyer''' from''' IIT Bombay''', signing off.
  
 
Thank you for joining.
 
Thank you for joining.
 
|-
 
|-
 
|}
 
|}

Latest revision as of 11:35, 15 January 2019

Visual Cue Narration
Slide Number 1

Title Slide

Welcome to this tutorial on Differentiation using GeoGebra.
Slide Number 2

Learning Objectives

In this tutorial, we will learn how to use GeoGebra to:

Understand Differentiation

Draw graphs of derivative of functions.

Slide Number 3

System Requirement

Here I am using:

Ubuntu Linux Operating System version 16.04

GeoGebra 5.0.481.0-d.

Slide Number 4

Pre-requisites

www.spoken-tutorial.org

To follow this tutorial, you should be familiar with:

GeoGebra interface

Differentiation

For relevant tutorials, please visit our website.

Slide Number 5

Differentiation: First Principles


f(x) = x2-x

f'(x) is derivative of f(x)

A (x, f(x)), B (x+j, f(x+j))

Differentiation: First Principles

Let us understand differentiation using first principles for the function f of x.

f of x is equal to x squared minus x


f prime x is the derivative of f of x.


Consider 2 points, A and B.

A is x comma f of x and B is x plus j comma f of x plus j

Show the GeoGebra window. I have opened the GeoGebra interface.
Type f(x)=x^2-x in the input bar >> Enter In the input bar, type the following line.

For the caret symbol, hold the Shift key down and press 6.

Press Enter.

Point to the equation in Algebra view.

Point to parabola in Graphics view.

Observe the equation and the parabolic graph of function f.
Click on Point on Object tool >> click on the parabola at (2,2).

Point to A at (2,2).

Click on Point tool and click on (3,6).

Clicking on the Point on Object tool, create point A at 2 comma 2 and B at 3 comma 6.
Click on Line tool >> click on points B and A. Click on Line tool and click on points B and A to draw line g.
Click on the Move tool.

Double click on the resulting line g and click on Object Properties.

Click on Color tab and select blue.

Click on Style tab and select dashed style.

As shown earlier in this series, make this line g blue and dashed.
Click on Tangents tool under Perpendicular Line tool. Under Perpendicular Line, click on Tangents.
Click on A >> click on the parabola. Click on A and then on the parabola.
Point to tangent h at point A to the parabola. This draws a tangent h at point A to the parabola.
Double click on tangent h and click on Object Properties.

Under Color tab, select red.

Close the Preferences box.

Let us make tangent h a red line.
Click on Point tool and click in Graphics view.

Point to point C.

Click on the Point tool and click anywhere in Graphics view.

This creates point C.

Double click on point C in Algebra view and change its coordinates to (x(B),y(A)).

Point to C.

In Algebra view, double-click on C and change its coordinates to the following.
Point to B and A. Now C has the same x coordinate as point B and the same y coordinate as point A.
Under Line, click on Segment and click on B and C, and then on A and C. Let us use the Segment tool to draw segments BC and AC.
Right-click on C >> Select Object Properties >> Color tab >> Purple

Click on Style tab >> select dashed line

Under Basic tab >> choose Name and Value >> Show Label check box.

Close the Preferences dialog box.

We will make AC and BC purple and dashed segments.
With Move highlighted, drag B towards A on the parabola. With Move highlighted, drag B towards A on the parabola.
Point to the value of j (length of AC) and lines g and h. Observe lines g and h and the value of j (length of AC).

As j approaches 0, points B and A begin to overlap.

Lines g and h also begin to overlap.

Point to line g, BC and AC. Slope of line g is the ratio of length of BC to length of AC.
Point to all the points on the parabola. Derivative of the parabola is the slope of the tangent at each point on the curve.
Point to text-box that appears in GeoGebra window.

As B approaches A, slope AB approaches slope of tangent at A.

As B approaches A on f of x, slope of AB approaches the slope of tangent at A.
Now let us look at the Algebra behind these concepts.
Slide Number 6

Differentiation: First Principles, the Algebra


f'(x) = lim_j→0 (length of Segment BC / length of Segment AC)

= lim_j→0 [(f(x+j) – f(x)]/[(x+j) – x]

Remember f(x) = x2-x, (x+j)2 = x2+2xj+j2

f'(x) = lim_j→0 [(x+j)2-(x+j)-(x2-x)]/(x+j-x)

Differentiation: First Principles, the Algebra

Slope of line AB equals the ratio of the lengths of BC to AC.

Line AB becomes the tangent at point A as distance j between A and B approaches 0.


BC is the difference between y coordinates, f of x plus j and f of x, for A and B.


AC is the difference between the x-coordinates, x plus j and x.

Let us rewrite f of x plus j and f of x in terms of x squared minus x.

We will expand the terms in the numerator.

Slide Number 7

The Algebra-Cont’d


f'(x) = lim_j→0 [x2+2xj+j2-x-j-x2+x]/j

= lim_j→0 [2xj+j2-j]/j = lim_j→0 [j(2x+j-1)]/j


= lim_j→0 [2x+j-1] = 2x-1

f'(x2-x) = 2x -1

After expanding the terms in the numerator, we will cancel out similar terms with opposite signs.


We will pull out j from the numerator and cancel it.


Note that as j approaches 0, j can be ignored. So that 2x plus j minus 1 approaches 2x minus 1.


As we know, derivative of x squared minus x is 2x minus 1.

Let us look at derivative graphs for some functions.
Slide Number 8

Differentiation of a Polynomial Function


Consider g(x)=5+12x-x3

Differentiation rules:

d(u±v)/dx = d(u)/dx ± d(v)/dx

d(5+12x-x3)/dx = d(5)/dx + d(12x)/dx - d(x3)/dx = 0 + 12 - 3x2 = -3x2 +12


For g(x)=5+12x-x3, g'(x) = -3x2 +12

Differentiation of a Polynomial Function

Consider g of x.

Derivative g prime x is the sum and difference of derivatives of the individual components.


g prime x is calculated by applying these rules.

Let us differentiate g of x in GeoGebra.
Open a new GeoGebra window. Open a new GeoGebra window.
Type g(x)=5+12x-x^3 in input bar >> Enter In the input bar, type the following line and press Enter.
Under Move Graphics View, click on Zoom Out.

Click in Graphics view until you see function g.

As shown earlier in the series, zoom out to see function g properly.
Right-click in Graphics view and select xAxis : yAxis option.

Select 1:5.

Right-click in Graphics view and select xAxis is to yAxis option.

Select 1 is to 5.

Under Move Graphics View, click on Zoom Out again.

Click in Graphics view to zoom out.

I will zoom out again.
Click on Point on Object tool and click on the curve to create point A.

Click on Tangent under Perpendicular Line.

Click on point A and the curve.

As shown earlier, draw point A on curve g and a tangent f at this point.
Click on Slope tool under Angle tool and on tangent line f. Under Angle, click on Slope and on tangent line f.
Point to slope of line f at A appearing as m value in Graphics view. Slope of tangent line f appears as m value in Graphics view.
Click on Point tool and in Graphics view to create point B.

Double click on point B in Algebra view and change coordinates to (x(A), m).

Point to points A and B and slope m of tangent line g.

Draw point B and change its coordinates to x A in parentheses comma m.
Right-click on point B >> select Trace On option. Right-click on B and select Trace On option
Click on Move tool and move point A on curve.

Observe the curve traced by point B.

With Move tool highlighted, move point A on curve.

Observe the curve traced by point B.

Let us check whether we have the correct derivative graph.
Type Deri in input bar >> select Derivative( <Function> ) >> Type g instead of highlighted <Function> >> press Enter In the input bar, type d e r i.


From the menu that appears, select Derivative Function option.

Type g to replace the highlighted word <Function>.


Press Enter.

Point to the equation in Algebra view.

Drag the boundary.

Note the equation of g prime x in Algebra view.

Drag the boundary to see it properly

Compare slides' calculations with equation of g'(x) in Algebra view. Compare the calculations in the previous slide with the equation of g prime x
Let us find the maxima and minima of the function g of x.
Point to derivative curve g'(x) above the x-axis and to g(x). Derivative curve g prime x remains above the x-axis (is positive) as long as g of x is increasing.
Point to derivative curve g'(x) below the x-axis and to g(x). g prime x remains below the x-axis (is negative) as long as g of x is decreasing.
Point to derivative curve g'(x) intersecting x-axis at x = -2 and x = 2. 2 and -2 are the values of x when g prime x equals 0.
Point to slope. Slope of the tangent at the corresponding point on g of x is 0.

Such points on g of x are maxima or minima.

Point to (-2,-11) and (2,21). Hence, for g of x, -2 comma -11 is the minimum and 2 comma 21 is the maximum.
Point to minimum of g(x) and x=-3 and x = -1. In GeoGebra, we can see that the minimum value of g of x lies between x equals -3 and x equals -1.
In the input bar, type Min.

From the menu that appears, select Min Function Start x-Value, End x-Value option.

Type g for Function.

Press Tab to go to the next argument.

Type -4 and -1 as Start and End x-Values.

Press Enter.

In the input bar, type Min.

From the menu that appears, select Min Function Start x-Value, End x-Value option.

Type g for Function.

Press Tab to go to the next argument.

Type -4 and -1 as Start and End x-Values.

Press Enter.

Point to minimum C in Graphics view and its co-ordinates in Algebra view. In Graphics view, we see the minimum on g of x.

Its co-ordinates are -2 comma -11 in Algebra view.

In the input bar, type Max.

From the menu that appears, select Max Function Start x-Value, End x-Value option.

Type g, 1 and 4 as the arguments.

Press Enter.

In the input bar, type Max.

From the menu that appears, select Max Function Start x-Value, End x-Value option.


Type g, 1 and 4 as the arguments.

Press Enter.

Point to maximum C in Graphics view and its co-ordinates in Algebra view. We see the maximum on g of x, 2 comma 21.
Finally, let us take a look at a practical application of differentiation.
Slide Number 9

A Practical Application of Differentiation


We have a 24 inches by 15 inches piece of cardboard

We have to convert it into a box

Squares have to be cut out from the four corners

What size squares should we cut out to get the maximum volume of the box?

We have a 24 inches by 15 inches piece of cardboard.

We have to convert it into a box.

Squares have to be cut from the four corners.

What size squares should we cut out to get the maximum volume of the box?

Slide Number 10

A Sketch of the Cardboard

Let’s draw the cardboard:

The volume function here is (24-2x)*(15-2x)*x cubic inches.

A Sketch of the Cardboard

Let us draw the cardboard:

This is the volume function here.

You could expand it into a cubic polynomial; but we will leave it as it is.

Open a new GeoGebra window. Open a new GeoGebra window.
Type (24-2 x) (15-2 x) x in the input bar >> Enter. In the input bar, type the following line and press Enter.
Drag the boundary to see the equation properly in Algebra view. Drag the boundary to see the equation properly in Algebra view.
Right-click in Graphics view and set xAxis : yAxis to 1:50.

Under Move Graphics View, click on Zoom Out.

Click in Graphics view to see the function properly.

Right-click in Graphics view and set xAxis is to yAxis to 1 is to 50.

Now, zoom out to see the function properly.

Point to the graph for this volume function in Graphics view.


Click in and drag the background to move Graphics view to see the maximum.

Observe the graph that is plotted for the volume function in Graphics view.

Drag the background to see the maximum.

Point to the maximum on top of the broad peak and to x = 0 and x = 7. Note that the maximum is on the top of this broad peak.
Point to both axes. The length of the square side is plotted along the x-axis.

Volume of the box is plotted along the y-axis.

In the input bar, type Max with capital M.


From the menu that appears, select Max Function Start x-Value End x-value.


Instead of highlighted Function, type f.


Press Tab to move and highlight Start x-Value and type 0.


Again, press Tab to move and highlight End x-Value and type 10.


Press Enter.

As before, let us find the maximum of this function.
Point to the maximum, A, in Graphics view and its coordinates in Algebra view. This maps the maximum, point A, on the curve.

Its coordinates 3 comma 486 appear in Algebra view.


Thus, we have to cut out 3 inch squares from all corners.

This will give the maximum possible volume of 486 cubic inches for the cardboard box.

Let us summarize.
Slide Number 11

Summary

In this tutorial, we have learnt how to use GeoGebra to:

Understand differentiation

Draw graphs of derivatives of functions

Slide Number 12

Assignment

Draw graphs of derivatives of the following functions in GeoGebra:

h(x)=ex

i(x)=ln(x)

j(x)=(5x3+3x-1)/(x-1)

Find the derivatives of these functions independently and compare with GeoGebra graphs.

As an assignment:

Draw graphs of derivatives of the following functions in GeoGebra.


Find the derivatives of these functions independently and compare with GeoGebra graphs.

Slide Number 13

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Slide Number 14

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Slide Number 15

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Slide Number 16

Acknowledgement

Spoken Tutorial Project is funded by NMEICT, MHRD, Government of India.

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This is Vidhya Iyer from IIT Bombay, signing off.

Thank you for joining.

Contributors and Content Editors

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