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UTFPR — Campus Pato Branco
Exercı́cios de Derivadas de Funções Reais de Variável Real
f (x + ∆x) − f (x)
f (x) − f (p)
ou f 0 (p) = lim
,
x→p
∆x→0
∆x
x−p
calcule a derivada das seguintes funções nos pontos dados:
1. Usando a definição de derivadas f 0 (x) = lim
(a) f (x) = 2x2 − 3x + 4, p = 2
(e) f (x) = 3 sin x, p = 2π e 0
√
3
(f ) v = √ − 2 t; t = 4
t
(g) f (x) = 5x − x2 , f 0 (−3) e f 0 (0)
9
(h) f (x) = x + , x = −3
x
3
,p=1
x2
√
(c) f (t) = 3 t, p = 8
(b) f (x) =
(d) g(x) = cos x, p =
π
2
2. Calcule a derivada das funções abaixo usando as propriedades adequadas:
(a) f (x) = 16x3 − 4x2 + 3
(d) f (t) = 2t − 1
(n) y = (x2 − 3x4 ) · (x5 − 1)
5t − 2
(o) g(t) =
1 + t + t2
(p) f (x) = (x2 + 3x + 3) · (x + 3)
(e) y = 8
(q) f (x) =
(b) f (x) = −5x3 + 21x2 − 3x + 4
(c) f (x) = 5
2x3
4x + 2
(r) y = (x + 2) · (x5 − 6x)
(f) y = 2x + 1
√
√
5
4
(g) y = x2 − x3 + x4
4
x2 − 4
x + 0, 5
√
1
(t) r = 2 · ( √ + θ)
θ
1
(u) f (x) = 2
(x − 1) · (x2 + x + 1)
1
(h) y = x 5 − x 6
(s) g(x) =
1000
(i) f (x) = 10100
5t − 1
(j) s(t) =
2t − 7
√
3
1
(k) f (x) = + 2 x − √
x
4 x
4
5
(l) f (r) = 2 + 3
r
r
(m) f (x) = (2x2 − 1) · (1 − 2x)
(v) v = (1 − t) · (1 − t2 )−1
√
1
(w) y = x + √
3
x4
3. Calcule a derivada das funções trigonométricas abaixo usando as regras de derivação:
(a) f (x) = tan x =
sin x
cos x
(e) y = x3 −
5
+ 2 sin x
(2x)3
3
(g) y = + 5 sin x
x
cotg x
(h) y =
1 + cotg x
1
(b) g(x) = sec x =
cos x
√
(c) f (x) = x · (2 sin x + x2 )
(d) h(θ) =
1
cos x
2
(f ) y =
π
sin θ − cos θ
2
4. Calcule a derivada das funções exponenciais e logarı́tmicas abaixo usando as regras de
derivação:
1
ex
cos x
y = ex · sin x
f (x) = x2 · ln x
f (x) = (x2 + 1) · ex
ex
y= x
2e + 1
(a) f (x) =
(f ) y = xex − ex
(b)
(c)
(d)
(g) y = x2 ex − xex
(e)
(h) y = 2ex
(i) y = e−t (t2 − 2t + 2)
dy
no ponto x = 1:
dx
3x + 2
· (x−5 + 1)
(c) y =
x
x+1
8
678
(d) y = (2x − x ) ·
x−1
5. Usando a regra do quociente e do produto, ache
(a) y =
2x − 1
x+3
(b) y =
4x + 1
x2 − 5
g(x) − g(2)
= 80.
x→2
x−2
6. Resolva e determine se é verdadeiro ou falso, se g(x) = x5 , então lim
7. Resolva e determine se é verdadeiro ou falso:
d
(10x ) = x10x−1
dx
d
1
(b)
(ln 10) =
dx
10
d
d
(tan2 x) =
(sec2 x)
dx
dx
d 2
(d)
|x + x| = |2x + 1|
dx
(a)
(c)
8. Derive utlizando as regras de derivação.
p
t · ln(t4 )
√
(t) y = sin(tan 1 + x3 )
(a) y = sin 4x
(s) y =
(b) y = cos 5x
(u) y = x · e3x
(c) y = e3x
√
x + 1(2 − x)5
(d) y =
(x + 3)7
(w) y = e−x · sin x
(e) y = sin t3
(x) y = e2t · sin 3t
(v) y = ex · cos 2x
2
(y) f (x) = e−x + ln(2x + 1)
et − e−t
(z) g(t) = t
e + e−t
cos 5x
(a1 ) y =
sin 2x
2
(b1 ) f (x) = (e−x + ex )3
(f ) g(t) = ln(2t + 1)
(g) x = esin t
(h) f (x) = cos(ex )
(i) y = (sin x + cos x)3
p
(j) y = (3x + 1)
s
x−1
3
(k) y =
x+1
(c1 ) y = t3 · e−3t
(d1 ) y = (sin 3x + cos 2x)3
√
(e1 ) y = x2 + e−x
(l) y = tan2 (sin θ)
(f1 ) y = x · ln(2x + 1)
(m) x = ln(t2 + 3t + 9)
(g1 ) y = [ln(x2 + 1)]
(n) f (x) = etan x
3
(h1 ) y = ln(sec x + tan x)
(o) y = sin(cos x)
(i1 ) f (x) = ln(x2 + 8x + 1)
√
(j1 ) f (x) = 6x + 2
(p) g(t) = (t2 + 3)4
(q) f (x) = cos(x2 + 3)
p
(r) y = (x + ex )
(k1 ) f (x) = x4 · e3x
(l1 ) f (x) = sin4 x
2
√
(e2 ) y = 2x x2 + 1
ex
(f2 ) y =
1 + x2
(g2 ) y = esin 2θ
(m1 ) f (x) = 5 tan 2x
(n1 ) f (x) = (2x3 − 3x) · (5 − x2 )3
3
(o1 ) f (x) = − √
3x − 5
(p1 ) y = ex
2 +x+1
(h2 ) y = emx cos nx
√
√
(i2 ) y = x cos x
(q1 ) y = sin 2x · cos x
(r1 ) y = (2x2 − 4x + 1)8
√
(s1 ) q = 2r − r2
3πt
3πt
(t1 ) s = sin
+ cos
2
2
√
(u1 ) h(x) = x tan(2 x) + 7
(x2 + 1)4
(2x + 1)3 (3x − 1)5
1
(k2 ) y =
sin(x − sin(x))
(j2 ) y =
(l2 ) y = ln(cossec 5x)
sec 2θ
1 + tan 2θ
(n2 ) y = ecx (c sin x − cos x)
(v1 ) r = sin(θ2 ) cos(2θ)
(m2 ) y =
(w1 ) y = (4x + 3)4 (x + 1)−3
(x1 ) y = x tan−1 (4x)
(o2 ) y = ln(x2 ex )
(y1 ) y = ecos x + cos(ex )
(p2 ) y = sec(1 + x2 )
(z1 ) y = cotg (3x2 + 5)
ex
(a2 ) y = −x
e +1
(b2 ) y = (x4 − 3x2 + 5)3
(q2 ) y = (1 − x−1 )−1
1
(r2 ) y = p
√
3
(x + x)
p √
(s2 ) y = sin x
1
(t2 ) y = ln(sin x) − sin2 x
2
(c2 ) y = cos(tan x)
3x − 2
(d2 ) y = √
2x + 1
9. Derive utilizando a derivada implı́cita:
(a) xy 4 + x2 y = x + 3y
(d) y = xey − y − 1
(b) x2 cos y + sin 2y = xy
(e) y 2 + x2 = 1
(c) sin(xy) = x2 − y
(f ) y 3 + yx2 = sen x + 3y 2 x
10. Encontre a derivada das seguintes funções:
(a) y = 8x
(g) y = (cos x)x
(b) y = 3cossec
(c) y = x(x
(h) y = x sinh x2
(x)
2 +1)
(i) y = ln(cosh 3x)
x2 +2x
(d) y = 7
(j) y = cosh−1 (sinh x)
(e) y = 3x ln x
(k) y = 10tan πθ
(f ) y = log5 (1 + 2x)
(l) y = x · tanh−1
11. Derive utilizando a derivada inversa:
(a) y = (arcsin 2x)2
(c) y =
√
(b) y = arctan(arcsin x)
√
x
(d) y = ln x
3
√
x
Respostas
1. (a) f 0 (2) = 5
(e) f 0 (2π) = 3
(b) f 0 (1) = −6
1
(c) f 0 (8) =
π 12
= −1
(d) g 0
2
11
16
0
(g) f (−3) = 11 e f 0 (0) = 5
(f ) v 0 (4) = −
(h) f 0 (−3) = 0
2. (a) f 0 (x) = 48x2 − 8x
(n) y 0 = x(−27x7 + 7x5 + 12x2 − 2)
(b) f 0 (x) = −15x2 + 42x − 3
(o) g 0 (t) =
(c) f 0 (x) = 0
(p) f 0 (x) = x2 + 4x + 4
(d) f 0 (x) = 2
(e) y 0 = 0
(q) f 0 (x) =
(f) y 0 = 2
2
3
(g) y 0 = √
− √
+ 4x3
5
5 x3 4 4 x
4
1
(h) f 0 (x) = √
− √
6
5
5 x 6 x5
0
(i) f (x) = 0
−33
(j) s0 (t) =
(2t − 7)2
3
1
1
(k) f 0 (x) = − 2 + √ + √
x
x 8x x
−8r − 15
(l) f 0 (r) =
r4
0
(m) f (x) = 2(−6x2 + 2x + 1)
x2 + x + 4
(s) g (x) =
(x + 0, 5)2
0
1
1
(t) r0 = √ − √
θ θ θ
(u) f 0 (x) =
(v) v 0 =
(e)
t2 − 2t − 1
(1 + t2 )2
(e) y 0 = 3x2 +
1
sin x
2
15
+ 2 cos x
8x4
3
(g) y 0 = 5 cos x − 2
x
1
(h) y 0 = −
2 sin x cos x + 1
(f ) y 0 = −
√
2 sin x + x2
√
+ 2 x(cos x + x)
2 x
π
cos θ + sin θ
2
ex (sin x + cos x)
f 0 (x) =
cos2 x
0
x
f (x) = e (sin x + cos x)
f 0 (x) = x(2 ln x + 1)
f 0 (x) = ex (x2 + 2x + 1)
ex
y0 =
(2ex + 1)2
(d) h0 (θ) =
(b)
(c)
(d)
−4x3 − 3x2 + 1
((x2 − 1)(x2 + x + 1))2
1
4
(w) y 0 = √ − 2 √
2 x 3x 3 x
(b) g 0 (t) = tan t · sec t
4. (a)
x2 (4x + 3)
(2x + 1)2
(r) y 0 = 3x5 + 5x4 − 6x − 6
3. (a) f 0 (x) = sec2 x
(c) f 0 (x) =
7 − 5t2 + 4t
(1 + t + t2 )2
(f ) y 0 = ex · x
(g) y 0 = ex (x2 + x − 1)
(h) y 0 = 2ex
(i) y 0 =
7
16
13
(b) y 0 (1) = −
8
(−t2 + 4t − 4)
et
(c) y 0 (1) = −29
5. (a) y 0 (1) =
(d) Descontı́nua em x = 1
6. Verdadeira
4
7.
5
(a) Falsa
(c) Verdadeira
(b) Falsa
(d) Falsa
8. (a) y 0 = 4 cos 4x
(l1 ) f 0 (x) = 4 sin3 x cos x
(m1 ) f 0 (x) = 10 sec2 2x
(b) y 0 = −5 sin 5x
(n1 ) f 0 (x) = (5 − x2 )2 [(6x2 − 3)(5 − x2 ) − 6x(2x3 − 3x)]
(c) y 0 = 3e3x
(b)
y0 =
2
√
1(2 − x)
5
x + 1(x + 3)7
−
√
5 x + 1(2 − x)4
(x + 3)7
−
√
7 x + 1(2 − x)5
(s1 ) q 0 = p
(k) y =
3
3x + 1
s
2
0
3(x + 1)2
3
·
(v1 ) r 0 = 2θ cos θ 2 cos 2θ − 2 sin θ 2 sin 2θ
x+1
2
(w1 ) y 0 =
x−1
(l) y 0 = 2 tan(sin θ) sec2 (sin θ) cos θ
(m) x0 =
(4x + 3)3 (4x + 7)
(x + 1)4
0
(x1 ) y = e sin ex − ecos x sin x
2t + 3
x
(y1 ) y 0 = 5 sec 5x
t2 + 3t + 9
(z1 ) y 0 = −6x · cossec2 (3x + 5)
(n) f 0 (x) = etan x sec2 x
(o) y 0 = − sin x cos(cos x)
0
2
(a2 ) y 0 =
3
(2e2x + e3x )
(ex + 1)2
(p) g (t) = 8t(t + 3)
0
(b2 ) y 0 = 6x(x4 − 3x2 + 5)2 (2x2 − 3)
2
(q) f (x) = −2x sin(x + 3)
(r) y 0 =
2
(c2 ) y 0 = − sin(tan x) sec2 x
1 + ex
√
x + ex
(d2 ) y 0 = √
(ln(t4 ) + 4)
(s) y = p
2 t ln(t4 )
0
(t) y 0 =
1−r
2r − r 2
3πx
3πx
3π
3π
(t1 ) s0 =
cos
sin
−
2
2
2
2
√
√
√
(u1 ) h0 (x) = tan(2 x) + x sec2 (2 x)
(i) y 0 = 3(sin x + cos x)2 (cos x − sin x)
√
(3x − 5)3
(r1 ) y 0 = 32(2x2 − 4x + 1)7 (x − 1)
2t + 1
(h) f 0 (x) = −ex sin ex
2
p
(q1 ) y 0 = 2 cos 2x cos x − sin 2x sin x
2
(g) x0 = esin t cos t
(j) y 0 =
2
2
(p1 ) y 0 = ex +x+1 (2x + 1)
(x + 3)8
(e) y 0 = 3t2 cos t3
(f ) g 0 (t) =
9
(o1 ) f 0 (x) =
3x2 cos(tan
1 + x3 ) sec2
p
2 1 + x3
p
p
(e2 ) y 0 =
1 + x3
(f2 ) y 0 =
(u) y 0 = e3x (1 + 3x)
3x + 5
2x + 1(2x + 1)
2(2x2 + 1)
p
x2 + 1
ex (1 + x2 − 2x)
(1 + x2 )2
(v) y 0 = ex (cos 2x − 2 sin 2x)
(g2 ) y = 2e
(w) y 0 = e−x (cos x − sin x)
(h2 ) y 0 = emx (m cos nx − n sin nx)
0
(x) y 0 = e−2t (3 cos 3t − 2 sin 3t)
2
0
(y) f (x) =
(z) g 0 (t) =
(a1 ) y 0 =
2x + 1
4e
(i2 ) y 0 =
−x2
− 2xe
(k2 ) y 0 =
(e2 t + 1)2
−5 sin 5x sin 2x − 2 cos 5x cos 2x
1
√
√
√
√ (cos x − x sin x)
x
2
cos x − cos(x − sin x)
sin2 (x − sin x)
(l2 ) y 0 = −5cotan 5x
sin2 2x
(m2 ) y 0 =
2
(b1 ) f 0 (x) = 3(e−x + ex )2 (−e−x + 2xex )
(c1 ) y 0 = 3t2 e−3t (1 − t)
0
2 sec 2θ(tan 2θ − 1)
(1 + tan2θ)2
(n2 ) y = e
2
(d1 ) y = 3(sin 3x + cos 2x) (3 cos 3x − 2 sin 2x)
(o2 ) y 0 =
2x − e−x
(e1 ) y = p
2 x2 + e−x
cx
(g1 ) y 0 =
x
(p2 ) y 0 = 2x sec(1 + x2 ) tan(1 + x2 )
2x
(q2 ) y 0 = −
(2x + 1)
1
(x − 1)2
√
1
2 x+1
(r2 ) y 0 = − √ p
√
3
6 x (x + x)4
√
cos x
(s2 ) y 0 = p
√
4 x sin x
6x[ln(x2 + 1)]2
x2 + 1
(h1 ) y 0 = sec x
(i1 ) y 0 =
(c2 sin x + sin x)
2+x
0
(f1 ) y 0 = ln(2x + 1) +
cos 2θ
(j2 ) y 0 = cotan 4x − 4x · cossec 4x
2t
0
2
sin 2θ
2x + 8
x2 + 8x + 1
(j1 ) f 0 (x) = √
(t2 ) y 0 = (cotan x − sin x cos x) =
3
6x + 2
(u2 ) y 0 = −
(k1 ) f 0 (x) = e3x x3 (4 + 3x)
9. (a) y 0 =
(b) y 0 =
(x2 + 1)3 (x2 + 56x + 9)
(2x + 1)4 (3x − 1)6
1 − y 4 − 2xy
4xy 3 + x2 − 3
(c) y 0 =
y − 2x cos y
2 cos 2y − x2 sin y − x
ey
(d) y =
2 − xey
0
6
cos3 x
(2x − y cos xy)
x cos xy + 1
sin x
(e) y 0 =
−x
y
cos x + 3y 2 − 2xy
(f ) y =
3y 2 + x2 − 6y
0
y0
y0
y0
y0
y0
= 8x ln(8)
= −3cossec x ln(3)cossec x · cotan x
2
2
= (x + 2 + 1)xx + xx +1 ln x · 2x
2
= 7x +2x ln(7)(2x + 2)
= 3x ln x ln 3(ln x + 1)
2
(f ) y 0 =
(1 + 2x) ln 5
0
(g) y = cosx x(ln(cos x) − x tan x)
10. (a)
(b)
(c)
(d)
(e)
(h) y 0 = sinh(x2 ) + 2x2 cosh(x2 )
3 sinh 3x
cosh 3x
sinh(sinh(x)) cosh x
(j) y 0 = −
cosh(sinh(x))2
(i) y 0 =
(k) y 0 = π10tan πx sec2 πx ln 10
√
√
√
2 tanh x − xsech2 x
0
(l) y =
√
2 tanh2 x
1
(c) y 0 = √
2 x
4 arcsin(2x)
11. (a) y 0 = √
1 − 4x2
√
cos x
0
√
(b) y = √
2 x(1 + sin2 x)
(d) y 0 =
1
x
Coletânea de exercı́cios elaborada pelos professores:
• Dra. Dayse Batistus;
• Msc. Ana Munaretto;
• Msc. Cristiane Pendeza;
• Msc. Adriano Delfino;
• Msc. Marieli Musial Tumelero
Digitação:
• 1a versão: Acadêmico Bruno Brito.
• Versão atual: Acadêmica Larissa Hagedorn Vieira.
Referência Bibliográfica:
ANTON, H., BIVENS, I. e DAVIS, S. Cálculo. vol. 1. Tradução: Claus I. Doering. 8 ed.
Porto Alegre: Bookman, 2007.
GUIDORIZZI, H. L. Um curso de cálculo, vol.1 e 2. 5a ed. LTC Editora, Rio de Janeiro,
RJ: 2002.
LEITHOLD, L. O cálculo com geometria analı́tica. Vol.1. 3a ed. São Paulo: Harbra,
1994.
LIMA, J. D. Apostila de Cálculo I. UTFPR, Pato Branco, 2008.
STEWART, James. Cálculo. Vol. 2. 6a ed. São Paulo: Pioneira Thomson Learning, 2009.
SWOKOWSKI, E. W. Cálculo com geometria analı́tica. Vol. 1. 2a ed. São Paulo: Makron
Books do Brasil,1994.
THOMAS, G. B. Cálculo. Vol. 1. 10a ed. São Paulo: Person, 2002.
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