Técnicas de derivação - Respostas

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1
Técnicas de derivação
RESPOSTAS
1. Definição de derivada: lim
h→0
f(x + h)− f(x)
h
(a) lim
h→0
1
3(x + h)
− 1
3x
h
= lim
h→0
(
1
3(x + h)
− 1
3x
)
· 1
h
= lim
h→0
3x− 3(x + h)
9x(x + h)
· 1
h
= lim
h→0
3x− 3x− 3h
9x(x + h)h
= lim
h→0
−3h
9x(x + h)h
= lim
h→0
−1
3x(x + h)
=
−1
3x(x + 0)
= − 1
3x2
(b) lim
h→0
1
2− (x + h)
− 1
2− x
h
= lim
h→0
[
1
2− (x + h)
− 1
2− x
]
· 1
h
= lim
h→0
2− x− [2− (x + h)]
[2− (x + h)][2− x]
· 1
h
= lim
h→0
2− x− 2 + x + h
(2− x− h)(2− x)h
= lim
h→0
h
(2− x− h)(2− x)h
= lim
h→0
1
(2− x− h)(2− x)
=
1
(2− x− 0)(2− x)
=
1
(2− x)2
2. (a)
d
dx
x5/3 =
5
3
x
5
3
−1 =
5
3
x2/3
(b)
d
dx
x1,4 = 1, 4x1,4−1 = 1, 4x0,4
(c)
d
dx
(
1√
x
)
=
d
dx
(
1
x1/2
)
=
d
dx
x−1/2
= −1
2
x(−
1
2
−1) =
x−3/2
2
= − 1
2x3/2
(d)
d
dx
x−4 = −4x−5
(e)
d
dx
(
1
4
√
x
)
=
d
dx
(
1
x1/4
)
=
d
dx
x−1/4
= −1
4
x−5/4
3. (a)
d
dx
(5t) = 5
d
dx
t1 = 5t1−1 = 5t0 = 5
d
dx
7 =
d
dx
(7t0) = 7 · 0 · t0−1 = 0
d
dx
(
3t5 + 2t3 − 5t + 7
)
=
d
dx
(3t5) +
d
dx
(2t3) − d
dx
(5t) +
d
dx
7
= 15t4 + 6t2 − 5
(b)
d
dx
[
1
2
(x3 + x−3)
]
=
1
2
· d
dx
(x3 + x−3)
=
1
2
·
(
d
dx
x3 +
d
dx
x−3
)
=
1
2
·
(
3x2 − 3x−4
)
=
1
2
· 3 ·
(
x2 − x−4
)
=
3
2
·
(
x2 − x−4
)
(c)
d
ds
(
s2 − 3s−2 + 5
√
s
)
=
d
ds
s2 − 3 d
ds
s−2 + 5
d
ds
s1/2
= 2s + 6s−3 +
5
2
s−1/2
4. (a)
d
du
(
u1/2
2
)
+
d
du
(
2
u1/2
)
=
1
2
· d
du
+ 2 · d
du
u−1/2 =
1
4
· u−1/2 − u−3/2
=
1
4
√
x
− 1
3
√
u2
(b)
d
dz
(
z3
3
)
− d
dz
(
z2
4
)
+
d
dz
(z
2
)
− d
dz
(
3
2
)
= z2 − z
2
+
1
2
5. (a)
d
dx
(x2 + 4x + lnx) = 2x + 4 +
1
x
(b)
d
dx
(10− lnx) = −1
x
(c)
d
dx
(
x lnx + 1
x
)
=
d
dx
(
x lnx
x
+
1
x
)
=
d
dx
(
lnx + x−1
)
=
1
x
− x−2 = 1
x
− 1
x2
=
x2 + x
x3
=
x(x + 1)
x3
=
x + 1
x2
6. (a)
d
dx
[
ln
(
5
x
)]
=
d
dx
(ln 5− lnx)
=
d
dx
ln 5− d
dx
lnx = −1
x
Como ln5 é uma constante, sua derivada é 0.
(b)
d
dx
lnx1/2 =
d
dx
(
1
2
lnx
)
=
1
2
· d
dx
lnx =
1
2x
(c)
d
dt
ln
(
2
t3
)
=
d
dt
ln 2t−3 =
d
dt
(ln 2 + ln t−3)
=
d
dt
ln 2 +
d
dt
ln t−3 = 0 +
d
dt
(−3 ln t)
= −3 d
dt
ln t = −3
t
7. (a)
d
dx
(
2ex − 1√
x
)
= 2 · d
dx
ex − d
dx
x1/2
= 2ex +
1
2
· x−3/2 = 2ex + 1
2x3/2
2
(b)
d
dx
(
ex + 1
2
)
=
d
dx
[
1
2
(ex + 1)
]
=
1
2
· d
dx
(ex + 1) =
1
2
· ex
(c)
d
dx
(elnx + ex + ln ex)
=
d
dx
elnx +
d
dx
ex +
d
dx
ln ex
=
d
dx
x +
d
dx
ex +
d
dx
x = 1 + ex + 1 = 2 + ex
(d)
d
dx
ex/3 =
1
3
ex/3
(e)
d
dt
(
1
et
)
=
d
dt
e−t = −e−t
(f)
d
dx
(
2x
3
+
2
3x
)
=
d
dx
(
2x
3
)
+
d
dx
(
2
3x
)
=
d
dx
(
1
3
· 2x
)
+
d
dx
(
2
3
· x−1
)
=
1
3
· d
dx
2x +
2
3
· d
dx
x−1
=
1
3
· ln 2 · 2x − 2
3
· x−2 = ln 2 · 2
x
3
− 2
3x2
(g)
d
dt
(
1
2t
)
=
d
dt
2−t = − ln 2 · 2−t = − ln 2
2t