DERIVADA DE FUNCIONES TRASCENDENTES
Enviado por SOL9 • 7 de Octubre de 2014 • 2.559 Palabras (11 Páginas) • 1.038 Visitas
Guadalajara, Jalisco 28 de agosto del 2014
Grupo: ER-ECDN-1402S-BI-001
Profesora: Eréndira Santos Viveros.
Alumno: Francisco Solís Mancilla
Materia: Calculo diferencial Unidad 3, Derivación
Actividad 2. Derivada de funciones trascendentes
Instrucciones: Resuelve los siguientes ejercicios determinando la derivada de las funciones o demostrando las expresiones que mencionan.
1. Calcula las siguientes derivadas:
a. d/dx ⌊(x²-1)/(x²+1)⌋^(1/2)
d/dx ⌊(x²-1)/(x²+1)⌋^(1/2) = 1/2 ⌊(x²-1)/(x²+1)⌋^((-1 )/2) *d/dx ⌊(x²-1)/(x²+1)⌋ = 1/2 ⌊(x²-1)/(x²+1)⌋^((-1 )/2) * ((x^2+1)(2x)-((x^2-1)(2x)))/(x^2+1)² =
1/2 ⌊(x²-1)/(x²+1)⌋^((-1 )/2)* (2x³+2x-2x^3+2x)/(x^2+1)²
d/dx ⌊(x²-1)/(x²+1)⌋^(1/2) = 1/2 ⌊(x²-1)/(x²+1)⌋^((-1 )/2)* 4x/(x^2+1)² = 4x/(2 √((x²-1 )/(x²+1 )) (x^2+1)²) = (2 x)/(√((x²-1 )/(x²+1 )) (x^2+1)²)
b. = 〖cos 〗[(X+4)/(X²-9)] * d/dx [(X+4)/(X²-9)] =
〖=cos 〗[(X+4)/(X²-9)]*((X^2-9)*d/dx (x+4) -((x-4)* d/dx 〖(x〗^2-9) ))/((x^2-9)²)
= cos [(X+4)/(X²-9)] * ((X^2-9) - (x-4)(2x))/((x^2-9)²) = cos [(X+4)/(X²-9)] *(x²-9-2x^2- 8x)/((x^2-9)²) =
= cos [(X+4)/(X²-9)]*(-x^2-8x-9)/(x²-9)² = cos [(X + 4)/(X²- 9)]*[- (x^2+ 8x + 9)/((x^2-9))] =
= cos [(X + 4)/(X²- 9)]*[- (x^2+ 8x + 9)/((x^2-9)²)] = [- (x^2+ 8x + 9)/((x^2-9)²)] cos [(X + 4)/(X²- 9)]
c. = ln〖sen (x^2)〗 = 1/(sen x^2 )(cosx²) (2x) =
2x ( cos〖(x^2)〗 ) 1/(sen x^2 ) = (2x cos x² )/(sen x^2 ) = 2x cotx²
d. = (√(x+1 ) d/dx(ln(x^2+1)+x^3-ln(x^2+1)+x³ d/dx(√(x+1)))/((√(x+1))²) =
= (√(x+1 ) 2x/(x²+1)+3x²-ln(x^2+1)+x³ 1/2 〖(x+1)〗^((-1)/2))/((√(x+1))²) = ((2x (√(x+1 )))/(x²+1)+3x²-ln(x^2+1)+ x³/(2√(x+1) ))/((√(x+1))²)
= ((2x (√(x+1 ))(x^2+1))/(x²+1)+3x²(x^2+1) (-ln(x^2+1)(x^2+1)+ x³/(2√(x+1) ) (x^2+1))/((√(x+1))²) =
= (2x √(x+1 )+3x^4+3x^2-(x^2+1) ln〖(x^2+1)+ x^3 (x^2+1)〗)/(x+1) =
= (2x √(x+1 )+3x^4+3x^2-(x^2+1) ln〖(x^2+1)+ x^5+x³〗)/(x+1) =
=(2x √(x+1 )+ 3x^2+ x³+3x^4+〖 x〗^5- (x^2+1)ln(x^2+1) )/(x+1) =
= (2x √(x+1 )+ 3x^2+ x³+3x^4+〖 x〗^5- x^2 ln(x^2+1)-ln(x^2+1))/(x+1)=
= (2x √(x+1 )+ x²(3-ln〖(x^2+1))〗+ x³+3x^4+〖 x〗^5-ln(x^2+1))/(x+1)
e. =
= [x³ e^4x.4 + e^4x .3x²] + ( e^(x^2 )-〖 sen〗x². 2x + cosx² e^(x^2 ). 2x) =
= [4x³ e^4x+ 3x²e^4x ] - 2x e^(x^2 ) 〖 sen〗x² + 2xe^(x^2 ) cosx² =
=4x³ e^4x + 3x²e^4x - 2x e^(x^2 ) senx² + 2xe^(x^2 ) cosx² =
= x² e^4x(4x + 3) - 2x e^(x^2 )(senx²- cosx²)
2. Demuestre dados se tiene que:
.
El seno hiperbólico y coseno hiperbólico se definen como:
〖sinh= 〗〖(e^x- e^(-x))/2〗 y 〖cosh= 〗〖(e^(x )+ e^(-x))/2〗
senhx coshy + coshx senhy = [(e^(x )- e^(-x))/2] [(e^(x )+ e^(-x))/2] + [(e^(x )+ e^(-x))/2] [(e^(x )- e^(-x))/2]
= 1/4 [e^(x+y )+ e^(x - y )- e^(x - y )- e^(-x - y ) ] + 1/4 [e^(x+y )+ e^(x - y )- e^(x - y )- e^(-x - y ) ]
= (e^(x + y )- e^(-x - y ) )/2 = senh〖(x+y〗)
3. Demuestre que dados con y se tiene que:
.
Si con y
sen x = 〖 cos 〗〖k π/2-x〗
cosx = -sen〖k π/2-x〗
tan〖x+y=((sen x)/cosx + (sen y)/cosy )/(1-(sen x)/cosx * (sen y)/cosy )〗 = ((sen x 〖 cos〗〖y +sen y cosx 〗)/cos〖x cosy 〗 )/((cosx cos〖y-senx sen y〗)/(cosx cosy )) = (sen x cos〖 y + senx cos〖 x〗 〗)/cos〖x cosy - senx sen y〗 =
(sen (x+y))/cos〖(x+y)〗 = tan x + y
4. Calcular los siguientes límites:
a. . = lim┬(x→4)〖(x³- 6x^2+ 5x + 12)/(x³- 3x^2- 6 x +8)〗
El teorema 3.1.14 sean f(x) y g(x) dos funciones derivables en un intervalo que contenga a x˳, supóngase que el lim┬(x→0)〖fx)= 〗 lim┬(x→0)〖g(x)=〗 0 y que gʹ (x̥) ≠ 0, existe entonces;
lim┬(x→0)〖(f(x))/(g (x))〗 Existe y lim┬(x→0)〖(f(x))/(g (x))〗 = (f ́ (x˳))/(g ́ (x ̥))
〖lim 〗┬(x→4)〖(x³-6x^2+5x+12)/(x³-3x^2-6x+8)〗 =
De acuerdo al teorema 3.1.14 se tiene f(x)= x³-6x^2+5x+12 y g(x)= x³-3x^2-6x+8
Son dos funciones derivables en 4, luego hay que observar que:
lim┬(x→4)〖x³-6x^2+5x+12 〗= (4)³ - 6(4)² + 5(4) + 12 = 64 - 96 +20 +12 = 0
〖lim 〗┬(x→4)〖x³-3x^2-6x+8 〗= (4)³ - 3(4)² - 6(4) + 8 = 64 – 48 – 24 + 8 = 0
El lim┬(x→0)〖fx)= 〗 lim┬(x→0)〖g(x)〗 0
Finalmente f ́ (x ̥) = 3x²-12x+5
g ́ (x ̥) = 3x²- 6x- 6
〖lim 〗┬(x→4)〖(3x²-12x+5)/(3x²-6x-6)〗 = ((3x²-12x+5)/3x²)/((3x²-6x-6)/3x²) = (1- (4 )/x+5/3x²)/(1- (2 )/x - 2/x²) = (1- (4 )/4+5/(3(4)²))/(1- (2 )/4 - 2/((4)²)) = (5/48)/(3/8) = 40/144
Si gʹ (x̥) ≠ 0, existe, por lo tanto el lim┬(x→0)〖(f(x))/(g (x))〗 = (f ́ (x˳))/(g ́ (x ̥))
〖lim 〗┬(x→4)〖(3x²-12x+5)/(3x²-6x-6)〗 = 10/36 = 5/18
...