Unit 5 – Derivatives of Trigonometric Functions
Topic 1: L’Hôpital’s Rule
Suppose we are trying to determine a limit such as the following:
lim
x 1
lnx
x1
We can see that this would give us a value of
0
- which would tell us we
0
need to try something else. Thinking about our previous strategies we
would try: factoring, rationalizing, finding a common denominator... none of
these work in this case.
lnx
, I would end up with
x x
Similarly if we were trying to evaluate this limit: lim

, but does that equal 1? Sometimes we can use reasoning to solve

0

these limits, but limits of the form or
are considered to be
0

indeterminate forms, and to solve these we need a new strategy!
L’Hôpital’s Rule Suppose f and g are differentiable and g  x  0near a .
Suppose that
limf  x 
and
xa
limf  x 
or that
limg  x 
xa
and
xa
limg  x  .
xa
Then
lim
xa
f  x
g  x

if the limit on the right side exists (or is  or  ).
Example 1:
lnx
.
x  1x  1
Find lim
SOLUTION:
Since limlnx  ln1 0 and lim x  1  0 we can apply L’Hôpital’s Rule.
x 1
Mrs. Lange
x 1
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
d
1
lnx
lnx
1
d
x
lim
 lim
 lim x  lim  1
x 1x  1 x 1 d
x 1 1
x 1x
x  1

dx
ex
Example 2: Calculate lim 2 .
x x
SOLUTION:
lnx
Example 3: Calculate lim 3 .
x 
x
SOLUTION:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Topic 2: Limits of Trigonometric Function (Section 7.1)
Before we can begin to find derivatives of trigonometric functions we first
have to start by looking at some very important limits (and as you recall the
derivative is just a limit anyway).
We’re going to start by looking at these limits:
limsin , limcos
 0
 0
Let’s look at the graphs of y  sin & y  cos to evaluate these limits.
y  sin
By looking at the graph; what is the value of limsin ?
 0
limsin 
 0
y  cos
By looking at the graph; what is the value of limcos ?
 0
limcos 
 0
The next very important limit that we need to evaluate is lim
 0
Since we don’t have the graph of y 
sin

sin

.
permanently engrained into our
brains, we’ll have to evaluate this limit using a table of values (this must be
done using RADIANS).
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
sinx
x
sinx
x
0.5 0.479
0.25 0.247
0.125 0.125
0.1
0.1
0.05
0.05
0.01
0.01
0.0000001 1E-07
From this table of values, we can see that lim
 0
sin


. We can use the
value of this limit to solve for other limits involving sin .
sin
.
 0 4
Example 1: Solve lim
SOLUTION:
We know lim
 0
sin

 1, so we want to be able to express our limit above in this
form somehow.
sin
1 sin
 lim 
  0 4
  0 4 
lim
 1  sin
  4lim
 
0


Example 2: Solve lim
 0
 1
 4

sin2
.
3
SOLUTION:
Let’s try something a little more complicated: (taken from text, p. 306 # 10)
lim
 0
cos  1

** Once we know how to take derivatives of trigonometric functions we will
be able to use
L’Hôpital’s Rule to solve these & other limits. **
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
For this type of problem we will need to use an identity. We need to use
one that involves both sinx and cosx - sin2 x  cos2 x  1. But we need to have
cos2 x in our limit in order to proceed.
SOLUTION:
lim
 0
cos  1


tan3x
x0 3tan2x
Another limit using trig identities: lim
As with other trig problems; we want to express this in terms of sinx and
cosx.
tan3x

x0 3tan2x
lim
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Topic 3: Derivatives of Sine & Cosine Functions
Recall: Angle Addition Identities:
cos(A B)  cosAcosB  sinAsinB
sin(A B)  sinAcosB  cosAsinB
d
sin(x  h)  sinx
sinx  lim
h0
dx
h
sinx cosh  sinhcosx  sinx
 lim
h0
h
sinx cosh  sinx  sinhcosx
 lim
h0
h
sinx  cosh  1
sinh cosx
 lim
 lim
h 0
h

0
h
h
 cosh  1  cosx lim sinh
 sinx lim
h 0
h 0 h
h
 sinx  0  cosx  1
d
sinx  cosx
dx
d
cos(x  h)  cosx
cosx  lim
h0
dx
h
Example 1:
Differentiate the following functions.
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
a. y  sin 2x  1
SOLUTIONS:
b. y  x3 cosx
y   2cos 2x  1
y   3x 2 cosx  x 3 sinx
y   cos 2x  1 2
y   3x 2 cosx  x 3   sinx 
Example 2:
Find y  given the following…
a. y  sin4  x2  1
SOLUTION:
c. y  2sinx
b. sinx  cosy  xy
SOLUTION:
Example 3:
Find the equation of the tangent line to y 
x

at the point where x  .
cos3x
3
SOLUTION:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Topic 4: Derivatives of Other Trig Functions
tanx 
sinx
cosx
cscx 
1
sinx
secx 
1
cosx
cot x 
1
cosx

tanx sinx
d  sinx  cosx  cosx   sinx   sinx  cos2 x  sin2 x
d
1


tanx  
 2  sec2 x

2
2
cos x
dx  cosx 
cos x
dx
cos x
d
cscx 
dx
d
secx 
dx
d
cot x 
dx
Example 1:
Differentiate f  x  tanx2 .
SOLUTION:


 
f   x   sec2  x2   2x   2x sec2 x2
Example 2:
Differentiate y  x2 cscx
SOLUTION:
y  2x cscx  x2   cscx cot x  2x cscx  x 2 cscx cot x  x cscx  2 x cot x
Example 3:
Find
dy
if tanx  secy  y  0.
dx
SOLUTION:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Example 4:
Sketch the curve y 
1
finding all intercepts, asymptotes, intervals of
cosx  sinx
increase/decrease, critical numbers, intervals of concavity and any points
 5 3 
of inflection on the domain x    ,  .
 4 4
SOLUTION:
Intercepts:
Asymptotes:
Intervals of Increase/Decrease & Critical Numbers:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Concavity & Points of Inflection:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Sketching the Graph:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Topic 5: Applications of Trig Functions & Their Derivatives
Example 1:
The position of a particle as it moves horizontally is described by the
function st   2sint  sin2t , 0 t   . If t is time in seconds, and s is
displacement in metres, find the max/min displacement.
SOLUTION:
Max/min displacement; that means we’re finding values of t when the
derivative is equal to zero.
s  2cost  2cos2t
s  2cost  2 2cos2 t  1


2cost  1 0


s  2 2cos t  2cost  cost  1
s  2 2cost  cost  1  1 cost  1 
s  2 2cos2 t  cost  1
2
cost  21  t 

3
max or min? FDT
cost  1 0
cost  1 t   max or min? FDT
s  2 2cost  1 cost  1
0 2 2cost  1 cost  1
0  2cost  1 cost  1
First Derivative Test:
s  t   2cost  2cos2t
s  0  2cos 0  2cos 2 0  2 2  0
 
 
 
s    2cos   2cos 2   0 2  0
 2
 2
 2
After t = π/3 s ’ changes from + to - ; so this is a maximum (based on FDT).
 3
s 
 2

 3
  2cos 2



 3
  2cos 2 2



  0 2  0

There is no sign change after t = π, so no max or min there (based on
FDT).
Maximum displacement occurs at t = π/3…
 2
 
 
s   2sin   sin
 3
 3
 3
3
3 3 3

  2 2  2  2  2.6m

Minimum displacement occurs at our endpoint (only because it’s an
endpoint):
s   2sin   sin 2   0m
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Example 2: Related rates!
The sides of a triangle have lengths 10 & 15 m. The angle between them
is increasing at a rate of

rad/s. How fast is the length of the third side
50
2
changing when the angle between the sides is
SOLUTION:
Step 1: Draw a picture.
known (has been given).
3
?
Step 2: Make a list of what is
Step 3: Identify what you are trying to find.
Step 4: Write an equation that relates what you know with what you are
trying to find.
Step 5: Solve!
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Topic 6: Inverse Trigonometric Functions
Now we are going to connect the information we saw earlier pertaining to
inverse functions with what we know about trigonometric functions.
y  sinx
Since this function fails
the horizontal line test (it
is not 1-1) we need to
restrict the domain to
  
 2 , 2 .


Information about Sine:


Domain: x |   x  ,x  
2
2

Range: y | 1 y  1,y 




Some points:   ,  1 ,  0, 0 &  , 1
 2

2 
Information about Arcsine (or Inverse Sine):
Domain:
Range:
Some points:
Now, the graph:
Graphing them
together:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
y  cosx
Since this function fails
the horizontal line test (it
is not 1-1) we need to
restrict the domain to
 0,  .
Information about Cosine:
Domain: x | 0 x   ,x  
Range: y | 1 y  1,y  
Some points:
 0, 1 ,
 
 2 , 0 &  ,  1


Information about Arc-cosine (or Inverse Cosine):
Domain:
Range:
Some points:
Now, the graph:
Graphing them
together:
y  tanx
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Since this function fails the horizontal line test (it is not 1-1) we need to
 
restrict the domain to   ,  .

2 2
Information about Tangent:
Arctangent (or Inverse Tangent):
Range:




x |   x  ,x  
2
2


y | y  
Asymptotes:
x
Some points:
 

 
  4 ,  1 ,  0, 0 &  4 , 1




Domain:

2
Information about
(vertical)
Now, the graph:
Graphing them together:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
The graph of this inverse function is different from the others – this one has
horizontal asymptotes. Because of this, the following limit statements can
be made:
lim tan1x 
x
lim tan1x 
x
Using Inverse Trig Functions:
Example 1:

Find the value of y  sin1 

2
.
2 
SOLUTION:
To find the answer we need to think critically about this situation. Inverse
 


sine has a range of   ,  .
 2 2
The question is asking us to find an angle that has a sine of 
Mrs. Lange
2
.
2
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
2
 
What angle between   ,  has a sine of  ?
2
 2 2

Thinking of the unit circle – it is y   .
4
Example 2:
Find the value of sin tan1 152  .
SOLUTION:
Example 3:
Find the value of cos1 cos56  .
SOLUTION:
Example 4:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Prove the following is true: tan1 53  sin1 53  tan1 2117 .
SOLUTION:
Example 5:
Express the following as an algebraic expression in terms of x: cos sin1x  .
SOLUTION:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Topic 7: Derivatives of Inverse Trig Functions
To derive these we need to use implicit differentiation and some trig
identities.
sin2   cos2   1
Recall:
tan2   1 sec2 
Derivative of Arcsine:
y  sin1x  siny  x
d
d
siny  x
dx
dx
dy
 cosy   1
dx
dy
1
1



dx cosy
1 sin2 y
d 1
sin x 
dx
1
1 x 2
1
1 x2
Derivative of Arccosine:
y  cos1x  cosy  x
d 1
cos x 
dx
Derivative of Arctangent:
y  tan1x  tany  x
d 1
tan x 
dx
Examples:
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
Differentiate the following functions:
1. y 
sin1 x
cos1 x
SOLUTION: We need to use the Quotient Rule!


cos1 x sin1 x  sin1 x cos1 x

y 

 cos1 x 
cos1x
y 
2. y 


2

cos1 x
   sin x   
1
1
1 x 2
 cos1 x 
2
   sin x       cos x  sin x 
1
1
1 x 2
 cos1 x
2
1
1
1 x 2
1 x 2
1
1 x 2
1
 cos1x
1
2
 cos x  sin x
1

1

1 x 2  cos1x
2
1
 cos x 
1 2
2
SOLUTION:
3. y  sin1 tan1x 
SOLUTION:
Example 4:
Find
dy
if y 2 sinx  tan1x  y
dx
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
SOLUTION:
Example 5:
Use what you know about curve sketching to graph the function:
y  sin1 cosx .
SOLUTION:
Domain:
Range:
Intercepts:
Asymptotes:
Increasing/Decreasing and Max/Min Values: (first derivative information)
Concavity/Points of Inflection: (second derivative information)
Mrs. Lange
Differential & Integral Calculus 120
Unit 5 – Derivatives of Trigonometric Functions
y  sin1 cosx
Mrs. Lange
Differential & Integral Calculus 120