Universidade Tecnológica Federal do Paraná 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. 7