MATH 12002 - CALCULUS I
§2.3, §2.4, and §2.5: Computing Derivatives (Part 2)
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
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Derivatives
10
f (x) = sin(x 2 + 2)
Use the Chain Rule:
f 0 (x) = [cos(x 2 + 2)] ·
d
dx (x
2
+ 2)
= [cos(x 2 + 2)] · 2x
= 2x cos(x 2 + 2).
D.L. White (Kent State University)
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Derivatives
11
f (x) = tan(2x − x 3 )
Use the Chain Rule:
f 0 (x) = [sec2 (2x − x 3 )] ·
d
dx (2x
− x 3)
= [sec2 (2x − x 3 )] · (2 − 3x 2 )
= (2 − 3x 2 ) sec2 (2x − x 3 ).
D.L. White (Kent State University)
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Derivatives
12
f (x) = x cos x
Use the Product Rule:
f 0 (x) =
d d
dx x cos x + x · dx cos x
= 1 · cos x + x(− sin x)
= cos x − x sin x.
D.L. White (Kent State University)
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Derivatives
13
f (x) = 4 sec7 (2 − 4x)
First rewrite the function as f (x) = 4[sec(2 − 4x)]7 , and then use the
Chain Rule twice:
d
f 0 (x) = 4 · 7[sec(2 − 4x)]6 · dx
sec(2 − 4x)
= 4 · 7[sec(2 − 4x)]6 · [sec(2 − 4x) tan(2 − 4x)] ·
d
dx (2
− 4x)
= 4 · 7[sec(2 − 4x)]6 · [sec(2 − 4x) tan(2 − 4x)] · (−4)
= 28[sec6 (2 − 4x)] · [sec(2 − 4x) tan(2 − 4x)] · (−4).
D.L. White (Kent State University)
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Derivatives
14
f (x) = (tan x)(sec x)
Use the Product Rule:
f 0 (x) =
d
dx
d
tan x (sec x) + (tan x) dx
sec x
= (sec2 x)(sec x) + (tan x)(sec x tan x).
D.L. White (Kent State University)
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Derivatives
15
f (x) = cos2 x − sin2 x
It may help to rewrite the function as f (x) = (cos x)2 − (sin x)2 .
Use the Chain Rule on each term:
d
d
cos x − 2(sin x)1 dx
sin x
f 0 (x) = 2(cos x)1 dx
= 2(cos x)(− sin x) − 2(sin x)(cos x)
= −4(sin x)(cos x).
Note: f (x) = cos2 x − sin2 x = 1 − 2 sin2 x = 2 cos2 x − 1.
Try computing the derivative using these other two forms of f (x).
D.L. White (Kent State University)
7 / 12
Derivatives
16
f (x) =
1 + tan x
1 − tan x
Use the Quotient Rule:
d
d
(1
+
tan
x)
(1
−
tan
x)
−
(1
+
tan
x)
(1
−
tan
x)
0
dx
dx
f (x) =
(1 − tan x)2
=
(sec2 x)(1 − tan x) − (1 + tan x)(− sec2 x)
.
(1 − tan x)2
D.L. White (Kent State University)
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Derivatives
17
f (x) =
sec x
x5
Use the Quotient Rule:
d
0
f (x) =
D.L. White (Kent State University)
dx
d 5
sec x · x 5 − (sec x) · dx
x
.
5
2
(x )
=
(sec x tan x) · x 5 − (sec x) · 5x 4
(x 5 )2
=
(sec x tan x) · x 5 − (sec x) · 5x 4
.
x 10
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Derivatives
18
f (x) =
sin(x 3 + 2)
cos(x 3 + 2)
Use the Quotient Rule, with the Chain Rule for the numerator and
denominator:
h
0
f (x) =
i
h
i
d
d
3
3
3
3
dx sin(x +2) cos(x +2)−[sin(x +2)] dx cos(x +2)
[cos(x 3 +2)]2
=
h
i
h
i
d
d
[cos(x 3 +2)]· dx (x 3 +2) cos(x 3 +2)−[sin(x 3 +2)][− sin(x 3 +2)]· dx (x 3 +2)
[cos(x 3 +2)]2
=
[cos(x 3 +2)]·3x 2 cos(x 3 +2)−[sin(x 3 +2)][− sin(x 3 +2)]·3x 2
.
[cos(x 3 +2)]2
3
sin(x +2)
3
Note: f (x) = cos(x
3 +2) = tan(x + 2).
Try computing the derivative using this other form of f (x).
D.L. White (Kent State University)
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Derivatives
19
f (x) = sin
√
x+
√
sin x
√
√
Observe that sin x = sin(x 1/2 ) and sin x = (sin x)1/2 .
Then use the Chain Rule on each term:
d√ 1
d
√
f 0 (x) = (cos x) · dx
x + 2 (sin x)−1/2 · dx
sin x
= (cos
D.L. White (Kent State University)
√
x) · 12 x −1/2 + 21 (sin x)−1/2 cos x.
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Derivatives
20
f (x) = sin(tan x + sec x)
Use the Chain Rule:
f 0 (x) = [cos(tan x + sec x)] ·
d
dx (tan x
+ sec x)
= [cos(tan x + sec x)](sec2 x + sec x tan x).
D.L. White (Kent State University)
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