Transcript Calculus 3.5 - Newmarket High School

3.5 Derivatives of Trig Functions
Explain why li m si n  0
limcos   1
 0
 0
Recall these two important limits:
lim
x 0
sin x
x
1
lim
and
cos x  1
x 0
x
0
Solve:
si n x
li m
x  0 2x
lim
x 0
sin2x
x
1 si n x
 li m
x 0 2
x
sin2x
 2 lim
x 0 2x
1
si n x 1

 li m
2
2 x 0 x
2
Go to Red text
P306 and do
#7-10, 13, 15,
17 and 22
Consider the function
y  sin  
We could make a graph of the slope:

slope



2
1
0
1
0
Now we connect the dots!

2
The resulting curve is a cosine curve.

1
0
d
sin  x   cos x
dx

f(x) = sin x
f’(x) = hlim
0
 lim
sin(x  h )  sin(x )
h
sinx cosh sinhcosx  sinx
h
h 0
 lim
sinx (cosh 1)  sinhcosx
h
h 0
 sinx lim
h 0
cosh 1
h
 cosx lim
h 0
sinh
h
 0  cos x

We can do the same thing for y  cos  
The resulting curve is a sine curve that has
been reflected about the x-axis.

slope



2
0
0
0

2
1

0
1
d
cos  x    sin x
dx


f(x) = cos x = si n(  x )
2

d 
( x)
f’(x) = cos(  x )
2
dx 2
 sinx (1)
  sinx

We can find the derivative of tangent x by using the
quotient rule.
cos 2 x  sin 2 x
=
cos 2 x
d
tan x
dx
d sin x
=
dx cos x
1
=
cos 2 x
cos x  cos x  sin x    sin x 
=
cos 2 x
2
sec
x
=
d
tan  x   sec 2 x
dx

Derivatives of the remaining trig functions can be
determined the same way.
d
sin x  cos x
dx
d
cot x   csc2 x
dx
d
cos x   sin x
dx
d
sec x  sec x  tan x
dx
d
tan x  sec2 x
dx
d
csc x   csc x  cot x
dx
p313 # 2abc, 3ab, 4a, 10 and p320 # 1abc, 3a, 13
