Chapter 2
Differentiation
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2-1 The Derivative and the Tangent Line Problem
2-2 Basic Differentiation Rules and Rates of Change
2-3 Product/Quotient Rule and Higher-Order Derivatives
2-4 Chain Rule
2-5 Implicit Differentiation
2-6 Related Rates (SKIP for now)
Day 1
2.2 Basic Differentiation and
Rates of Change
• Basic Differentiation Rules
• Constant Rule
• Power Rule
• Constant Multiple Rule
• Sum/Difference Rules
• Derivative of Sine and Cosine
(Skip until 2-3.)
• Rates of Change (Skip until Particle Motion)
http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx
Notations
There are many ways to denote the derivative of a function y = f(x). Some
common alternative notations for the derivative are
f '( x)  y ' 
dy df
d


f ( x)  D( f )( x)  Dx [ f ( x)]
dx dx dx
To indicate the value of a derivative at a specified number x = a, we use
the notation
f '(a) 
dy
df
d
|x  a 
|x  a 
f ( x) |x  a
dx
dx
dx
If the derivative of a function is its slope, then for
a constant function, the derivative must be zero
because it is a horizontal line.
We saw that if
y  x2 , y  2 x
y  x 3 , y  3x 2
3

f
x

4
x
f  x  x ,  
4
Recall the general binomial expansion for a positive integer n is
n 2
n
(
n

1
)
x
( x  h)n  x n  nx n 1 h 
h2  ...  h n
2
d n
n 1
x

nx


dx
Proof:
d n
( x  h) n  x n
x  lim
h0
dx
h
n(n  1) x n 2 2
x  nx h 
h  ...  h n  x n
d n
2
x  lim
h 0
dx
h
n 2
n
(
n

1
)
x
h(nx n 1 
h...  h n 1 )
d n
2
x  lim
h 0
dx
h
n
n 1
d n
x  nx n 1  0  0  ...  0  nx n 1
dx
Examples
Find the derivative of each of the following functions.
d
[ x11 ]
dx

f ( x)  e
d 
[ 2]
dx x
g ( x)  x
3
f ( x)  0
d 2
2
3
3
  [ x ]   (2 x )  2 x  3
dx
x
2
f ( x)  7 x  x  4
2
d 11
[ x ]  (11x10 )  11x10
dx
12
y  x3.7
g ( x)  x
2/3
2 1/3
2
g '( x)  x
 3
3
3 x
df
 14 x  12 x11
dx
y '  3.7 x 2.7
Harder Examples:
d  4 x3  2 x  7 


1.
dx 
x

4
3
4 x
2. Calculate f ' ( x) if f ( x)    3  .
4x x
8
Solutions:
d  4 x3  2 x  7 


dx 
x

4 x3  2 x  7 4 x3 2 x 7
Rewrite


  4 x 2  2  7 x 1
x
x
x
x
Differentiate.
d  4 x3  2 x  7  d

 
4 x 2  2  7 x 1
dx 
x
 dx





d
d
2  d 7 x 1
4x2 
dx
dx
dx
 4  2 x  0  7  1x 11

 8 x  7 x 2
7
 8x  2
x

Solutions:
3
4 x4
2. f ( x)    3 
4x x
8
Rewrite
Differentiate.
4
3 1
x
f ( x )   x  4 x 3 
4
8
3 2
1 3
4
f ' ( x)  x  12 x  x
4
2
3 12 1 3
f ' ( x)  2  4  x
4x
x
2
You Try…
Compute the derivative of each of the following
functions.
3 12
g ' (t )  t
2
1. g(t) = t3/2
1
1 1
2. r ( x) 
2  x 2
3x
3
1

0.9 0.9
3. y  3
 1  0.9 x 3
x
3
x
1 2  1
r ' ( x)   x  2
3
3x

dy
3
3
 0.3 x 
dx
10 x 4 / 3
4
You Try…
Compute the derivative of each of the following
functions.
3
5
v
5v 4 
1/ 4
v
4. u (v) 
 v
2
2
3
15v 2 1 3 / 4 15v 2
1
u ' (v ) 
 v

 3/ 4
2
4
2
4v
5 
 3 3
4 x 5  3x 2  5
  2 x   2 
5. y  
2
2 2x 

2x
3 5 2
3
 2 x   x
5
2
2
3
2 2
y '  6 x  5 x  6 x  3
x
You Try…
6. Calculate the slope and equation of the tangent
2
line to the curve y  2 x  1 at the point (1,3).
First find the derivative to calculate m.
dy
 4x
dx
Now, evaluate the derivative at x = 1 to find the slope at (1,3).
m  4 1  4
Use point slope form with slope, 4 and point (1,3).
y  y1  m x  x1 
y  3  4 x  1
Verify with a calculator.
Horizontal/Vertical Tangent
dy
0
Horizontal tangents occur when slope = zero, i.e.
dx
Vertical tangents occur when slope is undefined, i.e.
f ( x)  f (a )
f ( a  h)  f ( a )
lim
 
lim
 
OR h0
xa
xa
h
Find the horizontal tangent(s) of:
y  x  2x  2
4
2
dy
 4 x3  4 x
dx
Horizontal tangents occur when slope = zero.
4 x3  4 x  0
x3  x  0
x  x  1  0
2
x  x  1 x  1  0
x  0, 1, 1
Substituting the x values into the
original equation, we get:
y  2, y  1, y  1
The function is even, so we
only get two horizontal
tangents,
y = 1 and y = 2.
4
y  x4  2x2  2
3
y2
2
y 1
1
4
-2
-1
0
1
2
-1
3
-2
2
y  x4  2x2  2
dy
 4 x3  4 x
dx
1
-2
First derivative
(slope) is zero at:
x  0,1,1
-1
0
-1
-2
1
2
Find the vertical tangent(s)of f(x) = 3 x  2 .
lim f ( x)  f (a)
xa
xa
1
x  2  3 0  lim

lim
x 2 3
x 2
x 2
( x  2)2
3
The vertical tangent is at,
x = 2.
Check graphing calculator to verify.
You Try…
Find the horizontal/vertical tangent(s) of:
1. g ( x)  3 x 3  8 x 2  10
2. h( x)  x  x
3
Use a calculator to verify your answer.
Closure
Explain the difference between finding the
horizontal and vertical tangents of a graph.