MAC 2311 Calculus: Derivatives Basic Differentiation

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MAC 2311 CalculusDerivativesBasic
Differentiation

• Mrs. Kessler

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If the derivative of a function is its slope,
then for a constant function, the derivative must
be zero.
example
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Power Rule
This is part of a pattern.
examples
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Power Rule
Special Case? or just obvious
Dont forget it!
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Constant multiple rule
examples
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Sum and Difference rules
(Each term is treated separately)
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Derivative of ex

• Examples
• f(x) ex sin(x)
• f (x) ex cos(x)

f(x) 3ex x4 , f '(x) 3ex 4x3
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Basic Differentiation Formulas
Constant
Simple Power rule
Constant multiple rule
Sum and difference rule
Exponential
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Derivatives of Some Trigonometric Functions
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Derivative of Sine, Cosine

• Examples
• f(x) x2 sin(x)
• f(x) 2x cos(x)
• f(t) cos(t) 5t -2 , then f '(t)
  -sin (t) 10t -3

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Example
Find the horizontal tangents of
Plugging the x values into the original equation,
we get
(The function is even, so we only get two
horizontal tangents.)
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Find Equation of the Line Tangent to a Function

• Find the equation of the line tangent to the
  graph y x3- 2x 1 at (1, -2)
• Find slope- y 3x2-2
• Sub in x 1, m y 1
• y y1 m( x x1)
• y - -2 1(x-1)
• y x - 3

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Write the equation of the straight line
approximation
Use the Point-slope formula
Check on TI
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Find Equation of the Line Tangent to a Function

• Find the equation of the line tangent to the
  graph y x3- 2x 1 at (1, -2)
• Graph on TI 83/4
• 2nd Draw 5 ,ENTER
• Type 1(as in the point (1, -2)
• Enter
• y x - 3

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Motion/Velocity Example

• A dynamite blast blows a heavy rock straight up
  with a launch velocity of 160 ft/sec. Its height
  is given by
• s -16t2 160t.

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A dynamite blast blows a heavy rock straight up
with a launch velocity of 160 ft/sec. Its height
is given by s -16t2 160t.
a) How high does the rock go?
b) What is the velocity when the rock is 256 ft.
above the ground on the way up? On the way down?
c) What is the acceleration of the rock at any
time?
d) When does the rock hit the ground? At what
velocity?
The graphs of s and v as functions of time s is
largest when v ds/dt 0. The graph of s is not
the path of the rock It is a plot of height
versus time. The slope of the plot is the rocks
velocity graphed here as a straight line.
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A dynamite blast blows a heavy rock straight up
with a launch velocity of 160 ft/sec. Its height
is given by s -16t2 160t.
a) How high does the rock go?
Maximum height occurs when v 0.
s -16t2 160t
v s -32t 160
-32t 160 0
t 5 sec.
At t 5, s -16(5)2 160(5) 400 feet.
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A dynamite blast blows a heavy rock straight up
with a launch velocity of 160 ft/sec. Its height
is given by s -16t2 160t.
b) What is the velocity when the rock is 256 ft.
above the ground on the way up? On the way down?
-16t2 160t 256 -16t2 160t 2560 -16(t2 - 10t
16)0 -16(t 2) (t- 8) 0 t 2 or t 8
Set position 256
Find the times
v -32t 160 at t 2 v-32(2)160 96
ft/sec. at t 8 v-32(8)160 -96 ft/sec
Substitute the times into the velocity function
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A dynamite blast blows a heavy rock straight up
with a launch velocity of 160 ft/sec. Its height
is given by s -16t2 160t.
c) What is the acceleration of the rock at any
time?
s -16t2 160t
v s -32t 160
a v s -32ft/sec2
d) When does the rock hit the ground? At what
velocity?
s -16t2 160t 0 t 0 and t 10
Set position 0
v -32t 160 v -32(10) 160 -160 ft/sec.
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Derivative of y sin x
0 cos(x)1 cos (x)