Derivatives of Trig Functions

Suggested Practice Problems
Math 229, Calculus I
Differentiate the following:
Recall, for the function h(x) = f (g(x)), the chain rule is
h0 (x) = f 0 (g(x)) · g 0 (x)
Recall, for the function h(x) = f (x) · g(x), the product rule is
h0 (x) = f (x) · g 0 (x) + f 0 (x) · g(x)
f (x)
, the quotient rule is
g(x)
g(x) · f 0 (x) − f (x) · g 0 (x)
h0 (x) =
(g(x))2
Recall, for the function h(x) =
1.
d cos(x) sin4 (x)
dx
(a) Rewrite the function as cos(x) · (sin(x))4
(b) Now we differentiate using the product rule.
d d d
[cos(x)] · sin4 (x)
cos(x) sin4 (x) = cos(x) ·
(sin(x))4 +
dx
dx
dx
= cos(x) · 4 (sin(x))3 · cos(x) + [− sin(x)] · sin4 (x)
= 4 cos2 (x) sin3 (x) − sin5 (x)
= sin3 (x) 4 cos2 (x) − sin2 (x)
2.
d
[tan(x2 + sin(x))]
dx
Use the chain rule!
Recall
d
[tan(u)] = sec2 (u) · u0 . Note, u = x2 + sin(x).
dx
d d 2
tan(x2 + sin(x)) = sec2 (x2 + sin(x)) ·
x + sin(x)
dx
dx
= sec2 (x2 + sin(x)) · [2x + cos(x)]
d
1
3.
sec
dx
x2
(a) First, rewrite
1
as x−2 .
x2
(b) Now, we find
d
[sec(x−2 )]
dx
Note,
d
sec(u) = sec(u) tan(u) · u0 , where u = x−2 .
dx
d −2 d sec(x−2 ) = sec(x−2 ) tan(x−2 ) ·
x
dx
dx
= sec(x−2 ) tan(x−2 ) · −2x−3
= −2x−3 · sec(x−2 ) · tan(x−2 )
or
−2 sec(x−2 ) · tan(x−2 )
x3