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Section 3.3/3.4 - Techniques Of Differentiation/Derivatives Of
Trigonometric Functions
THEOREM: Let c be any real number and n be
any integer. If f and g are differentiable at x,
then so are f + g, f − g, f · g and
1. (cf )0 = cf 0
2. (f ± g)0 = f 0 ± g 0
3. (f g)0 = f 0g 0
0
f
f0
4.
= 0 , provided g(x) 6= 0
g
g
5. c0 = 0
6. (xn)0 = nxn−1
7. (sin x)0 = cos x
8. (cos x)0 = sin x
9. (tan x)0 = sec2 x
10. (cot x)0 = − csc2 x
11. (sec x)0 = sec x tan x
12. (csc x)0 = − csc x cot x
Section 3.3/3.4 - Techniques Of Differentiation/Derivatives Of
Trigonometric Functions
THEOREM: Let c be any real number and n be
any integer. If f and g are differentiable at x,
then so are f + g, f − g, f · g and
1. (cf )0 = cf 0
2. (f ± g)0 = f 0 ± g 0
3. (f g)0 = f 0g + f g 0
0
f
f 0g − f g 0
, provided g(x) 6= 0
4.
=
2
g
g
5. c0 = 0
6. (xn)0 = nxn−1
7. (sin x)0 = cos x
8. (cos x)0 = − sin x
9. (tan x)0 = sec2 x
10. (cot x)0 = − csc2 x
11. (sec x)0 = sec x tan x
12. (csc x)0 = − csc x cot x
(3x − 5)0 = (3x)0 − 50 = 3x0 − 50 = 3 · 1 − 0 = 3
√ 0 1/20 1 1/2−1 1 −1/2
( x) = x
= x
= x
2
2
1
√
x
0
=
1
x1/2
0
=
x−1/2
0
1 −1/2−1
=− x
2
1 −3/2
=− x
2
√ 0
√ 0
−8
−8
0
(−3x + 2 x) = (−3x ) + (2 x)
√ 0
−8
0
= −3(x ) + 2( x)
1 −1/2
−8−1
= −3(−8)x
+2 x
2
= 24x−9 + x−1/2
[(x3 + 7x2 − 8)(2x−3 + x−4)]0
= (x3 + 7x2 − 8)0(2x−3 + x−4)
+(x3 + 7x2 − 8)(2x−3 + x−4)0
= (3x2 + 14x)(2x−3 + x−4)
+(x3 + 7x2 − 8)(−6x−4 − 4x−5)
0
3x + 2
(x−5 + 1)
x+1
0
3x + 2
=
(x−5 + 1)
x+1
3x + 2
+
(x−5 + 1)0
x+1
(3x + 2)0(x + 1) − (3x + 2)(x + 1)0 −5
(x + 1)
=
2
(x + 1)
3x + 2
+
(x−5 + 1)0
x+1
3(x + 1) − (3x + 2) −5
=
(x + 1)
2
(x + 1)
3x + 2
(−5x−6)
+
x+1
(x3 sin x − 5 cos x)0
= (x3 sin x)0 − 5(cos x)0
= (x3)0 sin x + x3(sin x)0 − 5(cos x)0
= 3x2 sin x + x3 cos x + 5 sin x
sin x sec x
1 + x tan x
0
(sin x sec x)0 (1 + x tan x) − (sin x sec x)(1 + x tan x)0
=
(1 + x tan x)2
[(sin x)0 sec x + sin x(sec x)0 ](1 + x tan x) − (sin x sec x)(x0 tan x + x(tan x)0 )
=
(1 + x tan x)2
[cos x sec x + sin x sec x tan x](1 + x tan x) − (sin x sec x)(tan x + x sec2 x)
=
(1 + x tan x)2
THEOREM (The Chain Rule): If g is differentiable at x and f is differentiable at g(x), then the
composition f ◦g is differentiable at x. Moreover,
(f ◦ g)0(x) = f 0(g(x))g 0(x).
Alternatively, if
y = f (g(x)) and u = g(x)
then y = f (u) and
dy dy du
=
·
dx du dx
THEOREM: Let c be any real number and n be
any integer. If f, g and u are differentiable at x,
then so are f + g, f − g, f · g and
1. (cf )0 = cf 0
2. (f ± g)0 = f 0 ± g 0
3. (f g)0 = f 0g + f g 0
0
f 0g − f g 0
f
=
4.
, provided g(x) 6= 0
2
g
g
5. c0 = 0
6. (un)0 = nun−1 · u0
7. (sin u)0 = cos u · u0
8. (cos u)0 = − sin u · u0
9. (tan u)0 = sec2 u · u0
10. (cot u)0 = − csc2 u · u0
11. (sec u)0 = sec u tan u · u0
12. (csc u)0 = − csc u cot u · u0
[sin(3x)]0 = cos(3x) · (3x)0 = cos(3x) · 3
= 3 cos(3x)
[4 cos(x3)]0 = 4[cos(x3)]0
= −4 sin(x3) · (x3)0
= −12x2 sin(x3)
[(x2 − x + 1)23]0
= 23(x2 − x + 1)22 · (x2 − x + 1)0
= 23(2x − 1)(x2 − x + 1)22
√
[ x3 + csc x]0
= [(x3 + csc x)1/2]0
1 3
= (x + csc x)−1/2(x3 + csc x)0
2
1 3
= (x + csc x)−1/2(3x2 − csc x cot x)
2
1
x3 + 2x − 3
0
= [(x3 + 2x − 3)−1]0
= (−1)(x3 + 2x − 3)−2(x3 + 2x − 3)0
= −(3x2 + 2)(x3 + 2x − 3)−2
√
[sin( 1 + cos x)]0
√
√
= cos( 1 + cos x)[ 1 + cos x]0
√
1
= cos( 1 + cos x) (1 + cos x)−1/2(1 + cos x)0
2
√
1
= cos( 1 + cos x) (1 + cos x)−1/2(− sin x)
2