Applications-of-GeoGebra/C3/Differentiation-using-GeoGebra/English

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 OS 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 [[Image:]] f(x) = x2-x f'(x) is derivative of f(x) A (x, f(x)), B (x+j, f(x+j))	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 and press Enter. For the caret symbol, hold the Shift key down and press 6. f x in parentheses equals x caret 2 minus x
Point to the equation in Algebra view. Point to parabola in Graphics view.	The equation appears in the Algebra view and the function f is graphed as a parabola.
Point to parabola in Graphics view.	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).	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).	Create a point B at 3 comma 6.
Click on Line tool and click on points B and A.	Click on Line tool and click on points B and A.
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.	Let us make this line g blue and dashed.
Click on Tangents tool under Perpendicular Line tool.	Under Perpendicular Line, click on Tangents.
Click on A and then on the parabola.	Then 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.	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. Press Enter.
Point to C.	Now C has the same x coordinate as point B and the same y coordinate as point A.
	Let us draw segments BC and AC.
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. Then, click on A and C to draw AC. Note that BC and AC are called i and j in the order of their creation.
Right-click on AC >> Select Object Properties >> Color tab >> Purple Click on Style tab >> select dashed line	We will make AC and BC purple and dashed segments.
Under Basic tab >> choose Name and Value >> Show Label check box.	Under Basic tab, choose Name and Value from the dropdown menu next to the Show Label check box.
Close the Preferences dialog box.	Close the Preferences dialog box.
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 Band 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 slopes of tangents at all points on curve.
Point to the lines g and h.	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 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 7 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]	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 8 Differentiation: First Principles—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. 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. 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. 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.
Slide Number 10 Differentiation of a Polynomial Function Consider g(x)=5+12x-x3 d(5+12x-x3)/dx = d(5)/dx + d(12x)/dx - d(x3)/dx = 0 + 12 - 3x 2 = -3x 2 +12 For g(x)=5+12x-x3, g'(x) = -3x2 +12	Consider g of x equals 5 plus 12 x minus x cubed. 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.
	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. g x in parentheses equals 5 plus 12 x minus x caret 3
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.
Click on Point on Object tool and click on the curve to create point A.	Click on Point on Object tool and click on curve g of x to create point A.
Click on Tangent under Perpendicular Line. Click on point A and the curve.	Under Perpendicular Line, click on Tangent. Now click on point A and the curve g of x.
Point to tangent line f to the curve at point A.	This draws a tangent line f to the curve at point A.
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 line fat 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.
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.
Right-click on point B and select Trace On option.	Right-click on point B and select Trace On option
Click on Move tool and move point A on curve. Observe the curve traced by point B.	Click on Move tool and move point A on the curve. 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. From the menu that appears, select Derivative Function option. Type g to replace the highlighted word <Function>. Press Enter.
Point to the derivative graph.	This will confirm that you have the correct derivative graph.
Point to the equation of g prime x in Algebra view.	Note the equation of g prime x in Algebra view.
Drag the boundary to see it properly.	Drag the boundary to see it properly.
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).	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 the "peak" and "valley" on g of x.	Slope of the tangents at the corresponding points on g of x is 0. These points on g of x are maxima or a 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 with capital M. From the menu that appears, select Min 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 -3. Again, press Tab to move to and highlight End x-Value and type -1. Press Enter.	In the input bar, type Min with capital M. From the menu that appears, select Min 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 -3. Again, press Tab to move to and highlight End x-Value and type -1. 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 with capital M. 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. Again, press Tab to move to and highlight End x-Value and type 4. Press Enter.	In the input bar, type Max with capital M. 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. Again, press Tab to move to and highlight End x-Value and type 4. Press Enter.
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. Its co-ordinates are 2 comma 21 in Algebra view.
	Finally, let us take a look at a practical application of differentiation.
Slide Number 16 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 from the four corners 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 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 17 A Sketch of the Cardboard Let’s draw the cardboard: [[Image:]] 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. 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.
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.
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.
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.
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. Click in and drag the background to see Its coordinates 3 comma 486 appear in Algebra view. Thus, we have to cut out 3 inches squares from all corners. This will give the maximum possible volume of 486 cubic inches for the cardboard box.
	Let us summarize.
Slide Number 19 Summary	In this tutorial, we have learnt how to use GeoGebra to: Understand differentiation Draw graphs of derivatives of functions
Slide Number 16 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.
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	This is Vidhya Iyer from IIT Bombay, signing off. Thank you for joining.

Contributors and Content Editors

Madhurig, Snehalathak, Vidhya