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3.5 Derivatives of Trig Functions
2
0
2
Consider the function siny
We could make a graph of the slope: slope
1
0
1
0
1Now we connect the dots!
The resulting curve is a cosine curve. sin cosd
x xdx
What function does the red curve look like?
Derivative of y = sin x
USING LIMITS:
2
0
2
We can do the same thing for cosy slope
0
1
0
1
0The resulting curve is a sine curve that has been reflected about the x-axis.
cos sind
x xdx
We can find the derivative of tangent x by using the quotient rule.
tand
xdx
sin
cos
d x
dx x
2
cos cos sin sin
cos
x x x x
x
2 2
2
cos sin
cos
x x
x
2
1
cos x
2sec x
2tan secd
x xdx
Now use the quotient rule:
Derivatives of the remaining trig functions can be determined the same way.
sin cosd
x xdx
cos sind
x xdx
2tan secd
x xdx
2cot cscd
x xdx
sec sec tand
x x xdx
csc csc cotd
x x xdx
SAME Rules for Finding Derivatives
Simple Power rule 1n ndx nx
dx
Sum and difference rule ( ) ( )d d du v u v
dx dx dx
Constant multiple rule ( )d dcu c u
dx dx
Product rule ( ) ( )d d duv u v v u
dx dx dx
Quotient rule
2
( ) ( )d d
v u u vd u dx dxdx v v
Trig Identities
1cossin 22 xx
xx 22 sectan1
xx 22 csccot1
cossin22sin
22 sincos2cos
Don’t forget these!!!!
Example 1
• Find if
• We need to use the product rule to solve.
dx
dyxxy sin
xxxdx
dycossin)1(
xxxdx
dycossin
Example 2
• Find if
• We need to use the quotient rule to solve.
dx
dy
x
xy
cos1
sin
2)cos1(
)sin)((sin))(coscos1(
x
xxxx
dx
dy
xx
x
x
xxx
dx
dy
cos1
1
)cos1(
1cos
)cos1(
sincoscos22
22
Example 3
• Find if . )4/(// f xxf sec)(
xxxf tansec)(/
xxxxxxf tansectansecsec)( 2//
xxxxf sectansec)( 23//
)4/sec()4/(tan)4/(sec)4/( 23// f
23212)4/( 23// f
Find the derivatives21
( ) 5sin sec tan 7 32
f x x x x x x
21( ) 5cos sec tan sec tan (1) 14
2f x x x x x x x x
1 sin( )
cos
xf x
x x
2
( cos ) (1 sin ) (1 sin ) ( cos )( )
( cos )
d dx x x x x x
dx dxf xx x
2
( cos )(cos ) (1 sin )(1 sin )( )
( cos )
x x x x xf x
x x
2 2 2 2
2 2
( cos cos ) (1 sin ) cos cos 1 sin( )
( cos ) ( cos )
x x x x x x x xf x
x x x x
2
cos( )
( cos )
x xf x
x x
Simple Harmonic Motion
A weight hanging from a spring is stretched 5 units beyond its rest position (s = 0) and released at time t = 0 to bob up and down. Its position at any later time t is s = 5cos .
What is its velocity and acceleration at time t? Describe it’s motion.
Position :
Velocity :
Acceleration :
Find the slope at the given point:
1.) at the point (0, 1)
Find the slope at the given point:
1.) at the point (0, 1)
Find the slope at the given point:
1.) at the point (π, -π)
Find the slope at the given point:
1.) at the point
Find the derivative of each:1.) 2.)
)csc(csccotcotcsc' 2 xxxxxy
)csc(cotcsc' 22 xxxy
)csc(csccotcotcsc 2 xxxxxy
2)cos(sin
)sin(cos)cos)(sin1('
xx
xxxxxy
2)cos(sin
sincoscossin'
xx
xxxxxxy
Find the derivative of each:3.) 4.)
ttttg 2sectansec4)('
)sectan4(sec)(' ttttg
Find the derivative of each:5.) 6.)
Find the derivative of each:7.) 8.)
Find the derivative of each:9.) 10.)
Find: