Exercícios de Cálculo Diferencial

Prévia do material em texto

Correção páginas 59,60
1) Calcule: a) f(-1) e f(1/2) sendo 
f(-1) = -(-1)² +2.(-1) = -1-2=-3
f(1/2) = - (1/2)²+2.(1/2) = -1/4+1 = 3/4
g(0), g(2) e 
 sendo 
g(0) = 
		g(2) =
 sendo 
 e 
=
=
=
 sendo 
 e 
=
=
2) Sendo f(x) = 3, f(x) = 2x+1 e f(x) = -2x+3 e f(x) = x2 calcule se existir, em cada um dos casos, f´(0), utilizando a definição de derivada.
�� EMBED Equation.3 
3) Sendo f(x) = x2 + 1, calcule f´(1) utilizando a definição de derivada, encontre a equação da reta tangente ao gráfico de f(x) no ponto (1,2), e interprete geometricamente.
O coeficiente da reta tangente é dada pela derivada no ponto: Logo a = f’(1) = 2. E a equação é dada por y-y0 = a(x-x0) ( y-2 = 2(x-1)( y-2 = 2x-2( y = 2x
4) Encontre a taxa média de variação de cada função abaixo:
f(x) = 4-2x		com x0+h = 5 e x0 = 1	
 f(x) = -5x+7	com x0+h = 4 e x0 = 3	
f(x) = 3x2+x	com x0+h = 7 e x0 = 4	
f(x)= 3x+5		com x0+h = -1 e x0 = -2	
f(x)= 2x2-3x+1 	com x0+h= 3 e x0 =-2	
f(x) =12- x		com x0+h= 0 e x0 = -1	
 f(x) = 2x2 +4x 	com x0+h = 4 e x0 = 0	
f(x) = 3-6x		com x0+h= 6 e x0 = 5	
5) Calcule a derivada das seguintes funções:
f(x) = 1			( f’(x) = 0			
f(x) = x-2			 ( f’(x) = 1
f(x) = -x +1			 ( f’(x) = -1
f(x) =3x+12			 ( f’(x) = 3
f(x) = x3-4x			 ( f’(x) = 3x²-4
f(x)=
=x-2		( f’(x) = -2x-3 =- 
	
f(x) =
	=x1/2		( f’(x) = ½ x-1/2 =
 
f(x) = 
 (f’(x) = 
f(x) = 3x2+5x4+2x6		( f’(x) = 6x+20x³+12x5
f(x) = 2-x10			( f’(x) = -10x9
f(x) = 0,1.x10 +0,2x5		( f’(x) = 1x9+1x4
 f(x) = x2+5x-9 		( f’(x) =2x+5
f(x) = (2x+4).ex		( f’(x) = 2ex+(2x+4).ex
f(x) =
=x-1			( f’(x) = -1x-2 =- 
	
f(x)=
=x1/3		( f’(x) = 1/3 x-2/3 = 
p)f(x) = 
	( f ’(x) =
= 
= 
6) Calcule a derivada das funções nos pontos indicados:
f(x) = x2+1, 			no ponto x =3
f’(x) = 2x( f’(3) = 2.3 = 6
f(x) = 9x3-x, 			no ponto x = 1
f’(x) = 27x² -1 ( f’(1) = 27.1²-1 = 27-1 = 26
f(x) = -3x4+5x2, 			no ponto x = -2
f’(x) = -12x³ +10x ( f’(-2) = -12(-2)³ +10(-2) = 96-20 =76
f(x) = -x7, 				no ponto x = 1	
f’(x) = -7x6( f’(1) = -7.16 = -7
f(x) =
=x-1			no ponto x = 2
f’(x) = -1x-2 =- 
	( f’(2) = -
f(x) = (x5+2x3+5x).(2-x2-x4) 	 no ponto x = 0
f’(x) = (5x4+6x2+5).(2-x2-x4)+ (x5+2x3+5x).(-2x-4x3)
f’(0) = (5.04+6.02+5).(2-.02-04)+ (05+2.03+5.0).(-2.0-4.03) =5.2 = 10
f(x)=
=x-2			no ponto x=3
f’(x) = -2x-3 =- 
	( f’(3) = -
7) Calcule a derivada das seguintes funções simples:
a) y = -3				( y’=0	
b) y = -x3				(y’=-3x2			
c) y = 6x2				( y’=12x
d) y = 
				( y’=
e) y = 5x – 3x2 +4			(f’(x)=5-6x			
f) y = 7-x				(y’ =-1	
g) y = 
	(y’=
	
h) y = 6x 0,5				( y’ = 3x-0,5	
i) y = 
=x1/5+x1/3		(y’=
=
	
j) y = 5. ex+ 6. ln(x) +3. 2x + 6	(y’=5.ex + 6.
+3.2x.ln2		
k) y = 12x + x3			( y’=12x.ln12+3x2
l) y = 0				( y’=0	
m) y = 3-x6+x8 			( y’=-6x5+8x5
n) y = 4x+5x2+6x3+7x4 		(y’=4+10x+18x2+28x3
o) y =
		 	( y’ = 3/4
p) y = 0,2x+0,5x2-0,3		( y’=0,2+x
q) y = -0,6x			( y’ = -0,6
r) y = 
			(y’=
s) y = 7.ex + ln(x) - ln 2 		(y’=7ex+1/x
t) y = 
-3x+5= x1/5-3x+5		(y’=
-3 = 
u) y = 10x + 5. ln(x) + 3x+4		( y’=10x.ln10+5.1/x+3
v) y =5.3x				( y’ = 5.3x.ln3
Correção das Páginas 68 a 71
7) Sendo a, b, c e C constantes, ache as derivadas das seguintes funções: (constante, linear, constante vezes uma função, soma e diferença, potência e polinômios)
�
		(y’=0
		( y’ = 3
	( y’ = 5
		( y’ = 12x11
		( y’ = -12x-13
		( y’ = 4/3 x1/3
		( y’ = 24t²
	( y’ = 12t³ -4t
= x-4	( f’(x) = -4x-5 =- 
	
 	( f’(q) = 3q²
	(f’(x) = C.2x = 2Cx 
		( y’=2x+5
		( y’ = 18x²+8x-2
 		( y’ = -12x³-12x²-6
		( y’ = 8,4q -0,5
			( y’ = 2ax+b
 =z²+
		( y’ = 2z-
z-2 = 2z- 
 = 3t² +12t-1/2 –t -2	( y’ = 6t -6t-3/2 +2t-3 =
 = 3t² -5t1/2 +7t-1 	( y’ = 6t -5/2t-1/2-7t-2 = 
8)	Seja 
. Calcule as derivadas f’(0), f’(1), f’(2) e f’(-1). Verifique suas respostas graficamente.
F’(x) =2x ( f’(0) = 2.0 =0
F’(1) = 2.1 = 2
F’(2) = 2.2 = 4
F’(-1) = 2(-1) = -2
9)	Ache a oitava derivada de 
. Pense antes!
F(8) (x) = 0.
10)	Ache a equação da reta tangente ao gráfico de f em (1,1), onde f é dada por 
.
a = f’(1) = 6.1²-4.1 = 2
y – y0=a(x-x0) (y-1= 2(x-1)( y-1 =2x-2( y = 2x-2+1 = => y = 2x-1
11)	Ache a equação da reta tangente ao gráfico de 
 em t = 4. Esboce o gráfico de f(t) e da reta tangente nos mesmos eixos.
f’(t) = 6-2t ( f’(4) = 6-2.4 = -2
y= f(4) = 6.4-4² = 8
y – y0=a(x-x0) ( y-8 = -2(t-4)( y = -2t+8+8 ( y = -2t+16
12)	Sendo A, B e C constantes, ache as derivadas das seguintes funções: (+ exponencial e logarítmica)
�
		(y’ = 10t+4et
		( y’ =2ex +2x
		( f’(x) = 2x .ln2+2. 3x .ln3
		( y’ = 4.10x .ln10-3x²
		(y’ = 3-.4x .ln4
	(y’ = 3x-1.ln3-33/2 x-3/2 =
		( f’(x) = 3x² +3x.ln3
		( y’ = 5.5t .ln5 +6.6t .ln6
		(p’(t) = Cet
		( D’ = -1/p
			(R’ = 3.1/q = 3/q
		(y’=2t+5.1/t = 2t+5/t
		( y’ = A.et
	( f’(x) = Aex -2Bx
		( P’ = 9t² +2et
	( P’(t) = 3000.1,02t .ln1,02
	( P’(t) = 12,41. 0,94t .ln0,94
	( y’=5.2x.ln2 -5
		(R’(q) = 2q-2.1/q = 2q-2/q
	( y’ = 2x+4+3.1/x = 2x+4+3/x
 	( f’(t) = Aet +B.1/t =Aet +B/t
	�
13)	Ache as derivadas das seguintes funções: (+ regra da cadeia)
�
		( f’(x) = 99.(x+1)98.1= 99(x+1)98
		( R’= 4
.2q=8q. 
		( w’ = 100.(t2+1)99.2t = 200t.(t2+1)99
		( w’=100.(t3+1)99.3t2=300t2(t3=1)99
		( w’=3.(5r-6)2.5=15.(5r-6)2
			( f’(t) = 3.e3t
			( f’(t)=0,7 e0,7t
			( f’(t)=-4 e-4t
			( y’=1/2(s3+1)-½.3s2=3s2/2(s3+1)-½
				( w’= ½ s-1/2 
				( P’=-0,2 e-0,2t
				( w’=-6t
			( y’=
			( P’= 50.(-0,6) e-0,6t = -30. e-0,6t
			( P’=200.0,12 e0,12t= 24 e0,12t
		( y’=-6x+2.3e3x=-6x+6e3x
			( C’=12.3(3q2-5)2.6q=216q(3q2-5)2.
		( f’(x) =6.5e5x-2x
=30e5x-2x
				( y’= 5.5e5t+1=25 e5t+1
			( f’(x)=
			( f’(x)=
			( f’(x)=
			( f’(x)=
			( f’(t)= 5.
= 
			( g’(t)=
			( y’=
			( Q’=100.0,5(t2+5)-0,5.2t=100(t2+5)-0,5
			( y’=5+
			( y’ = 2.(5+ex)1.ex= 2ex(5+ex)
			(P’=0,5(1+lnx)-0,5.1/x�
14)	Se 
, ache f’(x) de dois modos: usando a regra do produto e antes efetuando a multiplicação para depois derivar. Compare os resultados!
Regra do produto(	f’(x) =2x(x³+5) +x²(3x²) = 2x4 +10x +3x4 = 5x4 +10x
Multiplicando antes(	f(x) = x5 +5x² (f’(x) = 5x4 +10x
15)	Se 
, ache f’(x) de dois modos: usando a regra do produto e 
antes efetuando a multiplicação para depois derivar. Compare os resultados!
Regra do produto(	f’(x) =2(3x-2) +(2x+1)3 = 6x-4+6x+3 = 12x-1
Multiplicando antes(	f(x) = 6x² -4x+3x-2 = 6x² -x-2 ( f’(x) = 12x -1
16)	Ache as derivadas das seguintes funções: (+ regra do produto e do quociente) 
		( f’(x) = 1.ex+x.ex= ex(1+x)
		( f’(t) =1.e-2t+t.(-2)e-2t=e-2t-2te-2t
			( y’=1.2x+x.2x.ln2
			( y’=5.
+5x.2x
=5
+10x2
		( y’=2t.(4t+1)3+t2.3(4t+1)2.4=2t+(4t+1)
			( y’=1.lnx+x.1/x = lnx+1
	( w’=(3t2+5).(t2-7t+2)+(t3+5t).(2t-7)
		( y’=2t.et+(t2+3).et
	( z’=3.(5t+2)+(3t+1).5=15t+6+15t+5=30t+11
�� EMBED Equation.3 	( y’=(3t2-14t)et+(t3-7t2+1).et
			( P’=2t.lnt+t2.1/t=2tlnt+t
		( f’(x)=
			( R’=3.e-q+3q.(-e-q)=3e-q -3qe-q
			(y’=1.
+ t.(-2t) 
=
-2t2
		( f’(z) = ½.z -½ .e-z +z ½ .(-e-z) 
	( g’(p)=1.ln(2p+1)+p.
=ln(2p+1)+
		( f’(t)=1.e5-2t+t(-2)e5-2t=e5-2t-2te5-2t
	( f’(w)=10w. 
+(5w2+3).2w. 
			( f ’(x)=
=
	( w’=
		( z’=
		( y’=
	( w’=
=
 =
			( y’=
17)	Sendo A, B e C constantes, ache as derivadas das seguintes funções: (+ funções periódicas)
		( y’ = 5.cos(x)
		( P’ = -sen(t)
		( y’ = 2t -5sen(t)
		(y’ = Acos(t)
	( y’ = 5cos(x) -5
	( R’(q) = 2q+2cos(q)
		( R’ = cos(5t).5
		( W’ = 4(-sen(t²)).2t = -8tsen(t²)
		( y’ = A.cos(Bt).B = AB.cos(Bt)
		( y’ = cos(x²).2x = 2x.cos(x²)
		( y’ = 2.(-sen(5t).5) =-10sen(5t)( y’ = 6.cos(2t).2 +3.(-sen(4t).4 = 12cos(2t) -12sen(4t)
		( f’(x) = cos(3x).3 = 3cos(3x)
		( z’ = -sen((4()
		( f’(x) = 2x.cos(x) +x² (-sen(x)) = 2xcos(x) –x²sen(x)
		( f’(x) = 2sen(3x) -2x.cos(3x).3 = 2sen(3x) -6xcos(3x)
		( f’(t) = 
		( f’(() = 
18) 	Sendo 
, sem construir o gráfico desta função, determine seu(s) ponto(s) de mínimo e de máximo, se existirem. Determine também o(s) intervalo(s) onde f(x) é crescente e o(s) intervalo(s) onde f(x) é decrescente, se existirem.
f’(x)= 2x-1
f’(x) >0( 2x-1>0 ( 2x>1 ( x >1/2 ( f é crescente
f’(x) <0( 2x-1<0 ( 2x<1 ( x<>1/2 ( f é decrescente				+
									 - 1/2
f’(x) = 0 ( 2x-1 = 0 ( 2x = 1( x = ½ é ponto crítico.
Pela ordem das setas (sinal da derivada) , vemos que ½ é ponto mínimo local. 
19) 	Sendo 
, sem construir o gráfico desta função, determine seu(s) ponto(s) de mínimo e de máximo, se existirem. Determine também o(s) intervalo(s) onde f(x) é crescente e o(s) intervalo(s) onde f(x) é decrescente, se existirem.
f’(x) = 2x² -4x-12 = 0
D = (-4)² - 4.2.(-12) = 16+96=112
X = (4±10,6)/4 ( x1 = 3,65 e x2 = -1,65 são pontos críticos.
		+		 -		+
			-1,65		3,65
Para x < -1,65 ou x > 3,65 temos que f’(x) >0, ou seja, em (-(,-1,65) ((3,65, +() f é crescente.
Para -1,65 < x < 3,65 temos que f’(x) <0, ou seja, em (-1,65, 3,65) f é decrescente.
Pela ordem dos sinais da derivada vemos que -1,65 é ponto mínimo local e 3,65 é ponto máximo local
20) 	Sendo 
, sem construir o gráfico desta função, determine seu(s) ponto(s) de mínimo e de máximo, se existirem. Determine também o(s) intervalo(s) onde f(x) é crescente e o(s) intervalo(s) onde f(x) é decrescente, se existirem.
F’(x) = 3(x-5)² .1 = 3(x-5)² = 0 ( x-5 =0 ( x = 5 é ponto crítico.
Como f’(x) = 3(x-5)² >0 para todo x real temos que f é sempre crescente. E não há ponto máximo nem ponto mínimo local.
21) Calcule a quinta derivada das funções:
f(x) = x10- 5x2+5		
f’(x) = 10x9 -10x
f’’(x) = 90x8 -10
f’’’(x) = 720x7
f(4) (x) = 5040x6
f(5) (x) = 30240x5 
f(x) = 3x7 - x5
f’(x) = 21x6 – 5x4
f’’(x) = 126x5 – 20x3
f’’’(x) = 630x4 – 60x²
f(4) (x) = 2520x3 – 120x
f(5) (x) = 7560x²– 120
f(x) =
 = x-1	
f’(x) = -1x-2
f’’(x) = 2x-3
f’’’(x) = -6x-4
f(4) (x) = 24x-5
f(5) (x) = 120x-6 = 120 / x6
f(x) = e2x	
f’(x) = e2x.2 =2. e2x.2
f’’(x) = 2.e2x.2 = 4 e2x
f’’’(x) = 4. e2x.2= 8e2x
f(4) (x) = 8. e2x.2 = 16 e2x
f(5) (x) = 16. e2x.2=32 e2x
	 
f(x) = 
= x-4
f’(x) = -4x-5
f’’(x) = 20x-6
f’’’(x) = -120x-7
f(4) (x) = 840x-8
f(5) (x) = 6720x-9 = 6720 / x9
22)
f(x) = e3-8x			(f’(x) = e3-8x.(-8) = -8e3-8x
f(x) = ln(2x-x4) 			(f’(x) = 2-4x³/2x-x4
f(x) = (5x-3x2)4 			( f’(x) = 4. (5x-3x2)3
f(x) = 
 		( 
f(x) = sen (4x-2)		(y’ =cos(4x-2) .4 = 4cos(4x-2)
f(x) = cos (x2-1)			( y’ = -sen(x²-1) .2x = -2xsen(x²-1)			
g) y = (2x-4)3			( y’ = 3.(2x-4)³.2=6(2x-4)³
h) y = (4 – 7x)7			( y’ = 7.(4-7x)6.(-7) = -49(4-7x)6
i) y = 2.e3x-1				( y’ = 2.3 e3x-1 = 6.e3x-1
j) y = 5.e2-x				( y’ = 5.(-1) e2-x = -5 e2-x
k) y = ln (3x-4)			( y’ = 
l) y = log 2 ( x+x2)			( y’ = 
m) y = 102x-3			( y’ = 2. 102x-3.ln10
n) y = 
=(3x+9)1/2		( y’ = ½ (3x+9)-1/2 .3 = 
o) y = (ln(x))3		( y’ = 3.(ln(x))².1/x = 3/x.(ln(x))².
p) f(x) = e5x-12		( f’(x) = e5x-12.5 = 5e5x-12
q) f(x) = ln(3x-x5)		(f’(x) = (3-5x4)/(3x-x5)
r) f(x) = (x-4x6)3		( f’(x) = 3.(x-4x6)² .(1-24x5)
s) f(x) = 
=
	( 
t) f(x) = sen (3x4-x)			( f’(x) = cos(3x4 –x).(12x³-1) 
u) f(x) = cos (4 -3x)			(f’(x) = -sen(4-3x)(-3) = 3sen(4-3x)			
v) y = (3x2+5)5			( y’ = 5.(3x²+5)4.6x =30x(3x²+5)4
x) y = (x3 –3x2)4			( y’ = 4(x³-3x²)3(3x²-6x)=(x³-3x²)(12x²-24x)
z) y = 
				( y’ = 2x
w) y = e5-2x				(y’ = -2.e5-2x
y) y = ln (x2-5x+1)			( y’ = 
aa) y = log (4-x2)			( y’ =
bb) y = 23x + 5.(3-x2)6 + e5x+2(y’=3.23xln2+5.6(3-x²)5(-2x)+5e5x+2=3.23xln2-60x(3-x²)5+5e5x+2
cc) y = 
=(x²-5x)1/3 	( y’ = 1/3 (x²-5x)-2/3.(2x-5) = 
dd) y = (e5x+3)4=e20x+12		(y’ = 20e20x+12
23) 
a) y = x2.ex			( y’ = u’.v+u.v’ = 2x.ex+x².ex
b) y = -x.ln(x)		( y’ = u’.v+u.v’ = -1.ln(x) +(-x).1/x = -ln(x) -1
c) y = x3. log(x)		( y’ = u’.v+u.v’ = 3x².log(x) +x³.1/x.ln10=3x².log(x) +x²/ln10
d) y = x.
=x.x1/2 = x3/2	( y’ = 3/2 .x1/2 =3/2 
	
e) y = 3x5.e4x+2		(y’ = u’.v+u.v’ =15x4.e4x+2+3x54.e4x+2 = 15x4.e4x+2+12x5.e4x+2
f) y =
			(y’= =	
g) y = 
 			( y’= 
 h) y = 
		( y’ = 
	
i) y = 
(	
	
j) y = 2x . x2			(y’ = u’.v+u.v’=2x.ln2.x²+2x.2x
k) y = x.(x+3)3		(y’=u’.v+u.v’ = 1(x+3)3+x.3(x+3)².1 = (x+3)³+3x(x+3)²
l) y = x2.(2x-1)4		( y’ = 2x(2x-1)4+x².4(2x-1)³.2=2x(2x-1)4+8x²(2x-1)³
m) y =
.x –1= xx1/2x-1 = x-1/2 	(y’ = -1/2 x-3/2
n) y = 3x2 e2-x. 		( y’=6x.e2-x+3x².(-1)e2-x =6xe2-x-3x²e2-x
o) y = 
	(
p) y = 
		(
q) y = 
	( y’ = 
r) y = 
	(	y’=
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