Gabarito L (6)

Prévia do material em texto

Ca´lculo I - Lista de Exerc´ıcios no¯ 6 - Gabarito - 1
o
¯ semestre/2016
1. (a) 7 (b) f(x) = 8x− 3 (c) f(x) =
−3
2x
√
x
(d) f(x) = −sen x
2. (a) 7 (b)
2
e5
3. (i) y
′
= x
2
3 (ii) y
′
= 21x6 − 12x2
(iii) y
′
=
−3
x
5
2
(iv) y
′
= 4
x2 + 2x− 4
(x2 + x+ 5)2
(v) y
′
=
−1
x ln2 x
(vi) y
′
=
x
2
+
−8
x3
(vii) y
′
= 2(10x4 + 14x3 + 3x2 − 8x− 7) (viii) y
′
= ln x
(ix) y
′
= 2xex + x2ex (x) y
′
= 2e
x(x+x lnx+1)
x
(xi) y
′
=
ex(x− 1)2
(x2 + 1)2
(xii) y
′
= −sen 2x+ cos2 x
(xiii) y
′
= 2xsen x+ cos x(x2 − 1) (xiv) y
′
= ex cos x − exsen x
(xv) y
′
= sec x tg x+ sec2 x (xvi) y
′
= sen x+ x cos x
(xvii) y
′
= − cos x(x3 + cos x) + (−3+ sen x)(−3x2 + sen x) (xviii) y
′
=
x+ ln(x) + 1
x2 ln2 x
(xix) y
′
= 4 cos x+ sen x+ sec xtg x (xxi) y
′
= ex(2+ sec2 x+ tg x)
(xx) y
′
= cos x+ 2x cos x− (1+ x2)sen x (xxii) y
′
= sec x(sec2 x+ tg 2x)
(xxiii) y
′
= −((1+x(1+x)cotgx)cosecx)
x2
(xxiv) y
′
=
1− ln x
x2
(xxv) y
′
= −cosec 2x+ 4 sec xtg x (xxvi) y
′
= 1
x ln 3
(xxvii) y
′
= 1
x lnpi (xxviii) y
′
= cos x+ sec2 x− xsen x
(xxix) y
′
= ex((1+ x) cos x− xsen x) (xxx) y
′
= x(cos x(3+ 2 ln x) − x(1+ ln x)sen x)
4. Para f(x) =
{
x+ 2, se x < 1
2, se x ≥ 1 , temos:
(i) Na˜o, o limite da f quando x tende a 1 na˜o existe.
(ii) Na˜o, pois se a derivada existisse f seria cont´ınua.
(iii) Na˜o e´ deriva´vel.
(iv)
1.
2.
3.
0
f
Para f(x) =
{
x2 − 2x+ 1, se x ≤ 1
−x2 + 2x− 1, se x > 1
, temos:
(i) Sim, pois 1 ∈ D(f) e lim
x→1 f(x) = 0 = f(1).
(ii) Sim, as derivadas laterais coincidem.
(iii) 0.
(iv)
1.0
f
Para f(x) =
{
−x− 1, se x ≤ 1
x2 − 3, se x > 1
, temos:
(i) Sim, p = 1 esta´ no domı´nio da func¸a˜o e o limite de f(x)
quando x tende a 1 existe e e´ igual a f(1).
(ii) Na˜o, as derivadas laterais na˜o coincidem.
(iii) Na˜o e´ deriva´vel.
(iv)
1.0
f
Instituto de Matema´tica Universidade Federal de Mato Grosso do Sul
5. (a) y
′
= lnx+ 1 se x > 0; y
′′
= 1
x
se x > 0; y
′′′
= −1
x2
se x > 0.
(b) y
′
= 2x
2
|x|
se x 6= 0; y ′′ = 2x3
|x|3
se x 6= 0; y ′′′ = 0 se x 6= 0
(c) f
′
(x) =
{
2x+ 3, se x ≤ 1
5, se x > 1
; f
′′
(x) =
{
2, se x < 1
0, se x > 1
; f
′′′
(x) =
{
0, se x < 1
0, se x > 1
6. (a) fn(x) = n! (b) fn(x) = (−1)
n+1(n−1)!
xn
.
7. f(27)(x) = −sen x
8. (a) y = 2x− 3
3/2.
−3.
0
f
g
(b) y = −4
25
x+ 13
25
13/4.
13/25.
0
f
a
(c) y = x+ 5
−5.
5.
0
f
g
A
(d) y = x; y = −1
e2
x+ 4
e2
;
4.
4/e2.
0
fa
b
A
B
9. y = 4x− 5; y = −x
4
+ 7
2
10. 13√
2
11. (i) y
′
= 4(−3+ 6x2)(7− 3x+ 2x3)3 (ii) y
′
= 4x√
(1+4x2)
(iii) y
′
= (1− 2x4)4(−6− 15x2 + 252x4 + 230x6) (iv) y
′
= (2x(−5+x
2)2(22+x2))
(4+x2)3
(v) y
′
= −2
(5−3x)(1/3)
(vi) y
′
= (x(−13+4x
2))√
(−4+x2)
(−1+x2)
)
(vii) y
′
= x
(25−x2)( 3
2
)
(viii) y
′
= −1
2
√
(5−x)
+ 3x
2
4(1+x3)(3/4)
(ix) y
′
= 1−10x+6x
2
2
√
(x(1−5x+2x2))
(x) y
′
= 1
2
√
( 2−x−3+x )(−3+x)
2
(xi) y
′
=
1
2
√
(x)
+2x
2
√
(
√
(x)+x2)
(xii) y
′
= 3
3x+4
(xiii) y
′
= 2
x
(xiv) y
′
= −2x
4−x2
(xv) y
′
= x
x2−5
(xvi) y
′
= 2x
−4+x2
(xvii) y
′
=
1
x
lnx (xviii) y
′
= 1+5x
2
2(x+x3)
(xix) y
′
= −1
2(6−5x+x2)
√
2+ln x−2
3−x
(xx) y
′
= 1
x ln|x|
(xxi) y
′
= 4e
(2t)
(1+e(2t))2
(xxii) y
′
= 5e5x
(xxiii) y
′
= 2xex
2
(xxiv) y
′
= 1√
x
e
√
x
(xxv) y
′
= −e−x(−2+ x)x (xxvi) y
′
= e−3x+x
3
(−1+ x2)
(xxvii) y
′
= 6(e−2x + x)(e−2x − x2)2 (xxviii) y
′
= −e2x(−2+ 3cotg (3x))cosec (3x)
(xxix) y
′
= −exsen (ex) (xxx) y
′
= −(cos cos x)sen x
(xxxi) y
′
= 2x cos(x2) (xxxii) y
′
= sen (2x)
Instituto de Matema´tica Universidade Federal de Mato Grosso do Sul
(xxxiii) y ′ = e−x(cos(x) − sen (x)) (xxxiv) y ′ = esen (t) cos(t)
(xxxv) y ′ = −6x cos(1− 3x2) sec2(sen (1− 3x2)) (xxxvi) y ′ =
−2x sec 1
−1+x2
tg 1
−1+x2
(−1+ x2)2
(xxxvii) y ′ =
−((−1+x) sec2(
√
(x
(1+x)
))
2
√
(x)(1+x)2
(xxxviii) y ′ = − cos( 1
1+x ) + 2(1+ x)sen (
1
1+x )
(xxxix) y ′ = −(x cos(
√
(1+x2))sen (sen (
√
(1+x2))))√
(1+x2)
(xl) y ′ = 2et
2
t cos(1+ et
2
)
(xli) y ′ = sec(1+ x2)(1+ 2x2tg (1+ x2)) (xlii) y ′ = sec x
(xliii) y ′ = 3 sec2(3x) (xliv) y ′ = 3x2 sec(x3)tg (x3)
(xlv) y ′ = −2xcosec 2(x2) (xlvi) y
′
=
(e−x(−(1+ 3x+ x2) cos(x) − x(1+ x)sen (x)))
(x2(1+ x)2)
(xlvii) y ′ = (e
2t(−3t+(1+5t+6t2) ln(1+3t)))
((1+3t) ln2(1+3t))
(xlviii) y ′ = (4(3 cos(3x) − 2sen (2x))(cos(2x) + sen (3x))3
(xlix) y ′ = e−x sec(x2)(−1+ 2xtg (x2)) (l) y ′ = −9x2 cos2(x3)sen (x3)
(li) y ′ = x2(4x sec2(4x) + 3tg (4x)) (lii) y ′ = x( (3x)(5+3x) + 2 log(5+ 3x))
(liii) y ′ = −6xcotg 2(x2)(x2 + cotg (x2))2 (liv) y ′ = x−1+sen (3x)(3x cos(3x) ln(x) + sen (3x))
(lv) y ′ = 1
x log(2) + 5
x log(5) (lvi) y ′ = 21+x
2
x ln(2) + 9x ln(9)
(lvii) y ′ = xx(cos(x) + (1+ ln(x))sen (x)) (lviii) y ′ =
((1+ 1
x
)x(−1+ (1+ x) ln(1+ 1
x
)))
(1+ x)
(lix) y ′ = (x
x(1+ln(x)))
(1+xx) (lx) y
′ = pix−1+pi + pix ln(pi)
12. (a) y ′ = ( 2
2x+1 +
2x
x2+3
+ 3x
2
x3−1
)(2x+ 1)(x2 + 3)(x3 − 1) (b) y ′ = (1
x
+ 1
x−1 +
1
x+2 −
1
x+1 )
x(x− 1)(x+ 2)
x+ 1
(c) y ′ = 1
2
( 2x
x2−1
− 2x
x2+1
)
√
x2 − 1
x2 + 1
(d) y
′
= xxxx
x
(ln2 x+ ln x+ 1
x
)
(e) y ′ = (ln x+ 1)xx2x
x
ln 2 (f) y ′ = (x2 + 1)cosx
[
−sen x ln(x2 + 1) +
2x cos x
x2 + 1
]
Instituto de Matema´tica Universidade Federal de Mato Grosso do Sul