1.4 the derivatives of some basic functions
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1.4 The Derivatives of Some Basic Functions. The derivative function represents the slope of the tangent at each point on the graph of the function. (where it exists) Derivative of y=x. Derivative of y=x 2. Slope of tangent at x =2x y’=2x. Derivative of y=x 2. y’=2x. - PowerPoint PPT PresentationTRANSCRIPT
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1.4 The Derivatives of Some Basic Functions
• The derivative function represents the slope of the tangent at each point on the graph of the function. (where it exists)
• Derivative of y=x.
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x y Slope of tangent at x
-3 9 -6
-2 4 -4
-1 1 -2
0 0 0
1 1 2
2 4 4
3 9 6
2
2
2
2
2
2
•Slope of tangent at x =2x
•y’=2x
Derivative of y=x2
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Derivative of y=x2
• The derivative of y=x2 is y’=2x
• See graph.
• See rate triangle
1
2=2x1
4=2x
The instantaneous rate of change of y with respect to x is 2x.
Visualize the point moving from left to right along the graph of y=x2 The slope along the tangent is always double the x-coordinate of the point.
y’=2x
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Derivative of y=x3
x y=x3 Slope of tangent at x
-3 -27 27
-2 -8 12
-1 -1 3
0 0 0
1 1 3
2 8 12
3 27 27
-15
-9
-3
3
9
15
•Slope of tangent at x is 3x2
•y’=3x2
6
6
6
6
6
9
4
1
0
1
4
9
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Derivative of y=x3
• The derivative of y=x3 is y’=3x2
• See graph.
• See rate triangle 13=3x2
The instantaneous rate of change of y with respect to x is 3x2.
Visualize the point moving from left to right along the graph of y=x3 . The slope along the tangent is always three time the square of the x-coordinate of the point.
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Graph of y=x3 and its derivative
y ‘ = 3x2
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Example 1• a) Find instantaneous rate of change of y=x2
at i) x=3 ii) x =-2
• b) Interpret graphically.
•Solution i)
•The derivative of y=x2 is y’=2x.
•When x=3, y’=2(3)=6
•y is increasing 6 times as fast as x is increasing when x=3.
m=6
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Example 1• a) Find instantaneous rate of change of y=x2
at ii) x =-2
• b) Interpret graphically.
•Solution ii)
•The derivative of y=x2 is y’=2x.
•When x=-2, y’=2(-2)=-4
•y is decreasing 4 times as fast as x is increasing when x=-2.
m=-4
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Example 2: Determine the equation of the tangent line.
• Determine the equation of the tangent to the graph of y=x3 at the point (-2,-8).
• Illustrate the results on a graph.•Solution
•The derivative of y=x3 is y’=3x2.
•When x=-2, the slope of the tangent line is y’=3(-2)2 = 12.
•The slope of the tangent is 12.
•The equation of the tangent line using y=m(x – p) + q is
•y=12(x+2)-8, or y=12x+16.
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Example 2: Graph• Graph the function and the tangent line.
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Derivative of y=mx+b
• y=x
• y=mx
• y=mx+b
• a) y’=1
• b) y’=m
• c) y’=m
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Vertical Translation Rule• When the graph of a function is translated
vertically, the derivative is not affected.• The derivative of y=f(x)+c is y’=f ’(x),
where c is any constant.• Example: Find the derivative of y=x2+4.• y’=2x.
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Vertical Translation RuleExample: Find the derivative of y=x2+4.• y’=2x.
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Vertical Stretch Rule
• When the graph of a function is expanded or compressed vertically, the graph of the derivative is also expanded or compressed vertically.
• The derivative of y=cf(x) is y’=cf’(x), where c is any constant.
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Vertical Stretch Rule• Example: Find the derivative of y=2x3
• f(x)=x3 , f’(x)=3x2
• y’=2f’(x) , so y’=2(3x2)• Or y’=6x2
•When the function is stretched by a factor of 2, the y-values for any x-value are doubled.
•That means when the slope is calculated at any point the slope will be doubled.
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Practice• Find the derivatives of the following
functions.
• a) y=x3+ 23
• b) y=-5x2
• c) y=4x3-7
• d) y=7x
• e) y=14 - x3
f(x)=x2 ;f ‘(x)=2x f(x)=x3 ;f ‘(x)=3x2
•a) y’=3x2
•b) y’=-10x
•c) y’=12x2
•d) y’=7
•e) y’=-3x2
If y = f(x) +c, then y ‘= f ‘(x) If y = cf(x), then y ‘= cf ‘(x)