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Transcript of MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you can...
![Page 1: MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you can derive a great deal of success from it. ” Chapter 2:](https://reader036.fdocuments.us/reader036/viewer/2022070400/56649f0e5503460f94c22208/html5/thumbnails/1.jpg)
MATH 31 REVIEWS
Chapter 2: Derivatives
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“ If you don't go off on a tangent while studying,
you can derive a great deal of success from it. ”
Chapter 2: Derivatives
![Page 3: MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you can derive a great deal of success from it. ” Chapter 2:](https://reader036.fdocuments.us/reader036/viewer/2022070400/56649f0e5503460f94c22208/html5/thumbnails/3.jpg)
1. Finding Derivatives from First Principles
Ex. If f(x) = 3x2 - 4x + 1 , find f (x) using limits.
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f(x) = 3x2 - 4x + 1
f (x) = lim f(x+h) - f(x)
h0 h
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f(x) = 3x2 - 4x + 1
f (x) = lim f(x+h) - f(x)
h0 h
= lim [3(x+h)2 - 4(x+h) + 1] - [3x2 - 4x + 1]
h0 h
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f(x) = 3x2 - 4x + 1
f (x) = lim f(x+h) - f(x)
h0 h
= lim [3(x+h)2 - 4(x+h) + 1] - [3x2 - 4x + 1]
h0 h
= lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
h0 h
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f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
h0 h
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f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
h0 h
= lim 6xh + 3h2 - 4h
h0 h
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f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
h0 h
= lim 6xh + 3h2 - 4h
h0 h
= lim h (6x + 3h - 4)
h0 h
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f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
h0 h
= lim 6xh + 3h2 - 4h
h0 h
= lim h (6x + 3h - 4)
h0 h
= lim (6x + 3h - 4) = 6x - 4
h0
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2. Sketching Derivative Functions
Remember, the derivative represents the tangent slope of a function.
Thus, the derivative function simply describes the slopes
of the original function.
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e.g.
Sketch the derivative function y = f (x)
y = f(x)
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y = f(x)Horizontal slopes ...
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y = f(x)Horizontal slopes ...
y = f (x)
... become x-intercepts (f = 0)
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y = f(x)Slope > 0
Slope > 0
y = f (x)
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y = f(x)Slope > 0
Slope > 0
y = f (x)Positive y-coordinates
Positive y-coordinates
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y = f(x)
Slope < 0
y = f (x)
Negative y-coordinates
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y = f(x)
y = f (x)
Degree = 3
Degree = 2
For polynomial functions, the degree decreases by 1
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3. Power Rule
Differentiate y = 5x3 - 7x + 9 - 4 + 1 - 2
x x2 x3
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y = 5x3 - 7x + 9 - 4 + 1 - 2
x x2 x3
y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3
Bring powers to the "top"
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y = 5x3 - 7x + 9 - 4 + 1 - 2
x x2 x3
y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3
y = 5 (3x2) - 7 + 0 - 4 (-1x-2) + -2x-3 - 2(-3x-4)
Ignore the coefficients
Differentiate the powers
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y = 5x3 - 7x + 9 - 4 + 1 - 2
x x2 x3
y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3
y = 5 (3x2) - 7 + 0 - 4 (-1x-2) + -2x-3 - 2(-3x-4)
y = 15x2 - 7 + 4x-2 - 2x-3 + 6x-4
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y = 5x3 - 7x + 9 - 4 + 1 - 2
x x2 x3
y = 5x3 - 7x + 9 - 4x-1 + x-2 - 2x-3
y = 5 (3x2) - 7 + 0 - 4 (-1x-2) + -2x-3 - 2(-3x-4)
y = 15x2 - 7 + 4x-2 - 2x-3 + 6x-4
y = 15x2 - 7 + 4 - 2 + 6
x2 x3 x4
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4. Product Rule
Differentiate y = (4x3 + 9x) (7x4 - 11x2 - 3) using
the product rule. No need to simplify the answer.
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y = (4x3 + 9x) (7x4 - 11x2 - 3)
y = (4x3 + 9x) (7x4 - 11x2 - 3) + (4x3 + 9x) (7x4 - 11x2 - 3)
Product Rule: f g + f g
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y = (4x3 + 9x) (7x4 - 11x2 - 3)
y = (4x3 + 9x) (7x4 - 11x2 - 3) + (4x3 + 9x) (7x4 - 11x2 - 3)
y = (12x2 + 9) (7x4 - 11x2 - 3) + (4x3 + 9x) (28x3 - 22x)
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5. Quotient Rule
Differentiate y = 5 - 2x using the quotient rule.
x2 + 4x
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y = 5 - 2x
x2 + 4x
y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x) (x2 + 4x)2
Quotient Rule: f g f g g2
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y = 5 - 2x
x2 + 4x
y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x) (x2 + 4x)2
y = (-2) (x2 + 4x) - (5 - 2x) (2x + 4)
(x2 + 4x)2
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y = 5 - 2x
x2 + 4x
y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x) (x2 + 4x)2
y = (-2) (x2 + 4x) - (5 - 2x) (2x + 4)
(x2 + 4x)2
y = -2x2 - 8x - (10x + 20 - 4x2 - 8x)
(x2 + 4x)2
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y = -2x2 - 8x - (10x + 20 - 4x2 - 8x)
(x2 + 4x)2
y = -2x2 - 8x - 10x - 20 + 4x2 + 8x
(x2 + 4x)2
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y = -2x2 - 8x - (10x + 20 - 4x2 - 8x)
(x2 + 4x)2
y = -2x2 - 8x - 10x - 20 + 4x2 + 8x
(x2 + 4x)2
y = 2x2 - 10x - 20
(x2 + 4x)2
or y = 2(x2 - 5x - 10)
x2 (x + 4)2
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6. Chain Rule
Differentiate y = 4 + 3x - 9x2
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y = 4 + 3x - 9x2
y = (4 + 3x - 9x2) 1/2
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y = 4 + 3x - 9x2
y = (4 + 3x - 9x2) 1/2
y = 1 (4 + 3x - 9x2) -1/2 d (4 + 3x - 9x2)
2 dx
Derivative of the "outside function"
Don't forget to find the derivative of the "inside function"
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y = 4 + 3x - 9x2
y = (4 + 3x - 9x2) 1/2
y = 1 (4 + 3x - 9x2) -1/2 d (4 + 3x - 9x2)
2 dx
y = 1 (4 + 3x - 9x2) -1/2 (3 - 18x)
2
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y = 1 (4 + 3x - 9x2) -1/2 (3 - 18x)
2
y = 3 - 18x
2 4 + 3x - 9x2
or
y = 3 (1 - 6x)
2 4 + 3x - 9x2
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7. Combination of Rules
Where does the function y = (2x - 7)4 (3x + 1)6
have a horizontal tangent?
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y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]
Use the product rule first
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y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)
Use the chain rule next
Don't forget to do the derivative of the "inside function"
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y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)
= 8 (2x - 7)3 (3x + 1)6 + 18 (2x - 7)4 (3x + 1)5
Put both terms into the same "order"
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y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)
= 8 (2x - 7)3 (3x + 1)6 + 18 (2x - 7)4 (3x + 1)5
= 2 (2x - 7)3 (3x + 1)5 [ 4(3x + 1) + 9(2x - 7) ]
Factor out the common factors
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y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1)6 + (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3 (2) (3x + 1)6 + (2x - 7)4 6(3x + 1)5 (3)
= 8 (2x - 7)3 (3x + 1)6 + 18 (2x - 7)4 (3x + 1)5
= 2 (2x - 7)3 (3x + 1)5 [ 4(3x + 1) + 9(2x - 7) ]
= 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
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y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
When does the function have a horizontal tangent?
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y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
The function has a horizontal tangent when y = 0
2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ] = 0
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y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
The function has a horizontal tangent when y = 0
2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ] = 0
when
2x - 7 = 0 or 3x + 1 = 0 or 30x - 32 = 0
x = 7 x = -1 x = 16
2 3 15
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8. Implicit Differentiation
Find y for the implicit relation x2y - 5x3 = y4 + 1
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7. Implicit Differentiation
Find y for the implicit relation x2y - 5x3 = y4 + 1
Remember, y is a function of x. So, you must treat it like
a separate function f(x).
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x2y - 5x3 = y4 + 1
(x2) y + x2 y - 15x2 = 4y3 y + 0
Product rule Chain rule
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x2y - 5x3 = y4 + 1
(x2) y + x2 y - 15x2 = 4y3 y + 0
2x y + x2 y - 15x2 = 4y3 y
![Page 51: MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you can derive a great deal of success from it. ” Chapter 2:](https://reader036.fdocuments.us/reader036/viewer/2022070400/56649f0e5503460f94c22208/html5/thumbnails/51.jpg)
x2y - 5x3 = y4 + 1
(x2) y + x2 y - 15x2 = 4y3 y + 0
2x y + x2 y - 15x2 = 4y3 y
x2 y - 4y3 y = 15x2 - 2xy
Get y terms on one side of the equations.
All others go on the other side.
![Page 52: MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you can derive a great deal of success from it. ” Chapter 2:](https://reader036.fdocuments.us/reader036/viewer/2022070400/56649f0e5503460f94c22208/html5/thumbnails/52.jpg)
x2y - 5x3 = y4 + 1
(x2) y + x2 y - 15x2 = 4y3 y + 0
2x y + x2 y - 15x2 = 4y3 y
x2 y - 4y3 y = 15x2 - 2xy
y (x2 - 4y3) = 15x2 - 2xy
Factor out the y
![Page 53: MATH 31 REVIEWS Chapter 2: Derivatives. “If you don't go off on a tangent while studying, you can derive a great deal of success from it. ” Chapter 2:](https://reader036.fdocuments.us/reader036/viewer/2022070400/56649f0e5503460f94c22208/html5/thumbnails/53.jpg)
x2y - 5x3 = y4 + 1
(x2) y + x2 y - 15x2 = 4y3 y + 0
2x y + x2 y - 15x2 = 4y3 y
x2 y - 4y3 y = 15x2 - 2xy
y (x2 - 4y3) = 15x2 - 2xy
y = 15x2 - 2xy
x2 - 4y3