Transcript B1.4 & B1.5 - Derivatives of Trigonometric Functions

B.4.6 - Derivatives of
Primary Trigonometric
Functions
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Lesson Objectives
• 1. Derive the derivatives of trigonometric
functions algebraically and graphically
• 2. Differentiate equations involving
trigonometric functions
• 3. Apply sinusoidal functions and their
derivatives to real world problems
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Fast Five
• 1. State the value of sin(/4), tan(/6), cos(/3),
sin(/2), cos(3/2)
• 2. Solve the equation sin(2x) - 1 = 0
• 3. Expand sin(x + h)
• 4. State the value of sin-1(0.5), cos-1(√3/2)
• 5. Explain how to find the value of the limx->0
(cosx - 1)/x
• 6. Explain how to find the value of the limx->0
sin(x)/x
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Explore
• Explain how to determine the value of the
following limits numerically & graphically
• Now do it
• Now use the TI-89 Calculus menu to verify
the value of your limits
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(A) Derivative of the Sine
Function - Graphically
•
•
•
•
•
•
•
We will predict the what the derivative
function of f(x) = sin(x) looks like from
our curve sketching ideas:
We will simply sketch 2 cycles
(i) we see a maximum at /2 and -3 /2
 derivative must have x-intercepts
(ii) we see intervals of increase on (2,-3/2), (-/2, /2), (3/2,2) 
derivative must increase on this
intervals
(iii) the opposite is true of intervals of
decrease
(iv) intervals of concave up are (-,0)
and ( ,2)  so derivative must
increase on these domains
(v) the opposite is true for intervals of
concave up
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(A) Derivative of the Sine
Function - Graphically
•
•
•
•
•
•
•
•
We will predict the what the derivative
function of f(x) = sin(x) looks like from
our curve sketching ideas:
We will simply sketch 2 cycles
(i) we see a maximum at /2 and -3 /2
 derivative must have x-intercepts
(ii) we see intervals of increase on (2,-3/2), (-/2, /2), (3/2,2) 
derivative must increase on this
intervals
(iii) the opposite is true of intervals of
decrease
(iv) intervals of concave up are (-,0)
and ( ,2)  so derivative must
increase on these domains
(v) the opposite is true for intervals of
concave up
So the derivative function must look
like  the cosine function!!
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(B) Derivative of Sine Function Algebraically
•
We will go back to our limit concepts for determining the derivative of y =
sin(x) algebraically
f (x)  lim
h0
d
sin(x) 
dx
d
sin(x) 
dx
d
sin(x) 
dx
d
sin(x) 
dx
d
sin(x) 
dx
d
sin(x) 
dx
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f (x  h) f (x)
h
sin(x  h) sin(x)
lim
h0
h
sin(x)cos(h) sin(h)cos(x) sin(x)
lim
h0
h
sin(x)[cos(h) 1)] sin(h)cos(x)
lim
h0
h
sin(x)[cos(h) 1]
sin(h)cos(x)
lim
 lim
h0
h0
h
h
cos(h) 1
sin(h)
lim(sin(x)) lim
 lim
 lim cos(x)
h0
h0
h0
h0
h
h
cos(h) 1
sin(h)
sin(x) lim
 cos(x) lim
h0
h0
h
h
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(B) Derivative of Sine Function Algebraically
• So we come across 2 special trigonometric limits:
•
sin(h)
lim
and
h0
h
cos(h) 1
lim
h 0
h
• So what do these limits equal?


• We will introduce
a new theorem called a Squeeze (or
sandwich) theorem  if we that our limit in question lies
between two known values, then we can somehow
“squeeze” the value of the limit by adjusting/manipulating
our two known values
• So our known values will be areas of sectors and triangles
 sector DCB, triangle ACB, and sector ACB
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(C) Applying “Squeeze
Theorem” to Trig. Limits
1
A = (cos(x), sin(x))
0.8
0.6
D
0.4
0.2
-1.5
-1
C
-0.5
0.5
E = (1,0)
B = (cos(x), 0)
1
1.5
-0.2
-0.4
-0.6
-0.8
-1
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(C) Applying “Squeeze
Theorem” to Trig. Limits
•
•
We have sector DCB and sector ACB “squeezing” the triangle ACB
So the area of the triangle ACB should be “squeezed between” the
area of the two sectors
1
A = (cos(x), sin(x))
0.8
0.6
D
0.4
0.2
-1.5
-1
C
-0.5
0.5
E = (1,0)
B = (cos(x), 10)
1.5
-0.2
-0.4
-0.6
-0.8
-1
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(C) Applying “Squeeze
Theorem” to Trig. Limits
•
Working with our area relationships (make h =  )
1 (OB) 2 ( )  1 (OB)(OA)  1 (OC) 2 ( )
2
2
2
1   cos2 ( )  1 sin( ) cos( )  1   (1) 2
2
2
2
 cos2 ( )  sin( ) cos( )  
 cos2 ( ) sin( ) cos( )



 cos( )
 cos( )
 cos( )
sin( )
1
cos( ) 


cos( )
•
We can “squeeze or sandwich” our ratio of sin(h) / h between cos(h)
and 1/cos(h)
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(C) Applying “Squeeze
Theorem” to Trig. Limits
•
Now, let’s apply the squeeze theorem as we take our limits as h
0+ (and since sin(h) has even symmetry, the LHL as h 0- )
sin(h)
1
 lim
h0
h 0
h  0 cos(h )
h
sin(h)
1  lim
1
h 0
h
sin(h)
 lim
1
h 0
h
lim cos(h)  lim
•
Follow the link to Visual Calculus - Trig Limits of sin(h)/h to see
their development of this fundamental trig limit
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(C) Applying “Squeeze
Theorem” to Trig. Limits
•
Now what about (cos(h) – 1) / h and its limit  we will treat this algebraically
cos(h)  1
h 0
h
cos(h)  1cos(h)  1
 lim
h 0
hcos(h)  1
lim
cos2 ( h)  1
 lim
h  0 hcos(h)  1
 sin 2 ( h)
 lim
h  0 hcos(h)  1
sin(h)
sin(h)
 1 lim
 lim
h 0
h  0 cos(h )  1
h
 0 
 1 1 

 11
0
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(D) Fundamental Trig. Limits:
Graphic and Numeric Verification
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•
•
•
•
•
•
•
•
•
•
•
•
•
x
-0.05000
-0.04167
-0.03333
-0.02500
-0.01667
-0.00833
0.00000
0.00833
0.01667
0.02500
0.03333
0.04167
0.05000
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y
0.99958
0.99971
0.99981
0.99990
0.99995
0.99999
undefined
0.99999
0.99995
0.99990
0.99981
0.99971
0.99958
14
(D) Derivative of Sine
Function
•
Since we have our two fundamental trig limits, we can now go back and
algebraically verify our graphic “estimate” of the derivative of the sine
function:
sin(h)
1
h 0
h
cos(h)  1
lim
0
h 0
h
d
cos(h)  1
sin(h)
sin(x)   sin(x)  lim
 cos(x)  lim
h 0
h 0
dx
h
h
d
sin(x)   sin(x)  0  cos(x) 1
dx
d
sin(x)   cos(x)
dx
lim
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(E) Derivative of the Cosine
Function
• Knowing the derivative of the sine function,
we can develop the formula for the cosine
function
• First, consider the graphic approach as we
did previously
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(E) Derivative of the Cosine
Function
•
•
•
•
•
•
•
•
We will predict the what the derivative
function of f(x) = cos(x) looks like from our
curve sketching ideas:
We will simply sketch 2 cycles
(i) we see a maximum at 0, -2 & 2 
derivative must have x-intercepts
(ii) we see intervals of increase on (-,0), (,
2)  derivative must increase on this
intervals
(iii) the opposite is true of intervals of
decrease
(iv) intervals of concave up are (-3/2,-/2)
and (/2 ,3/2)  so derivative must increase
on these domains
(v) the opposite is true for intervals of
concave up
So the derivative function must look like 
some variation of the sine function!!
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(E) Derivative of the Cosine
Function
•
•
•
•
•
•
•
•
We will predict the what the derivative
function of f(x) = cos(x) looks like from our
curve sketching ideas:
We will simply sketch 2 cycles
(i) we see a maximum at 0, -2 & 2 
derivative must have x-intercepts
(ii) we see intervals of increase on (-,0), (,
2)  derivative must increase on this
intervals
(iii) the opposite is true of intervals of
decrease
(iv) intervals of concave up are (-3/2,-/2)
and (/2 ,3/2)  so derivative must increase
on these domains
(v) the opposite is true for intervals of
concave up
So the derivative function must look like 
the negative sine function!!
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(E) Derivative of the Cosine
Function
• Let’s set it up algebraically:
d
d  

cos(x)   sin  x  
dx
dx   2

d
cos(x)  d  sin   x    d    x 
dx
2


  dx  2

d  x  
2

d


cos(x)  cos  x   (1)
dx
2

d
cos(x)  sin(x)  1   sin(x)
dx
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(F) Derivative of the Tangent
Function - Graphically
• So we will go through our
curve analysis again
• f(x) is constantly
increasing within its
domain
• f(x) has no max/min
points
• f(x) changes concavity
from con down to con up
at 0,+
• f(x) has asymptotes at +3
• /2, +/2
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(F) Derivative of the Tangent
Function - Graphically
• So we will go through our curve
analysis again:
• F(x) is constantly increasing within
its domain  f `(x) should be
positive within its domain
• F(x) has no max/min points  f
‘(x) should not have roots
• F(x) changes concavity from con
down to con up at 0,+  f ‘(x)
changes from decrease to increase
and will have a min
• F(x) has asymptotes at +3 
• /2, +/2  derivative should have
asymptotes at the same points
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(F) Derivative of the Tangent
Function - Algebraically
•
We will use the fact that tan(x) = sin(x)/cos(x) to find the derivative of tan(x)
d
tan(x)  d  sin(x) 
dx
dx  cos(x) 
d
d

sin(x)  cos(x)  cos(x)  sin(x)
d
dx
tan(x)  dx
dx
cos(x) 2
d
tan(x)  cos(x)  cos(x)  2 sin(x)  sin(x)
dx
cos x
d
cos2 x  sin 2 x
tan(x) 
dx
cos2 x
d
tan(x)  12  sec2 x
dx
cos x
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(H) Examples
• 1. Differentiate:
(i) y = sin(3x), (ii) y = cos(x + 2), (iii) y = sin(kx + d)
• 2. Differentiate:
(i) y = sin(x3), (ii) y = cos3x,
(iii) y = sin3(x2 - 1), (iv) y = x2cosx
• 3. Find the equation of the tangent line to y =
sin(x)/cos(2x) at x = pi/6
• 4. Find where f(x) = x - cosx is increasing and concave
up
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(G) Internet Links
• Calculus I (Math 2413) - Derivatives - Derivatives of Trig
Functions from Paul Dawkins
• Visual Calculus - Derivative of Trigonometric Functions
from UTK
• Differentiation of Trigonometry Functions - Online
Questions and Solutions from UC Davis
• The Derivative of the Sine from IEC - Applet
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(H) Homework
• Stewart, 1989, Chap 7.2, Q1-5,11
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