DERIVADA DE FUNCIONES TRASCENDENTES

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 )(cos⁡x²) (2x) =

2x ( cos⁡〖(x^2)〗 ) 1/(sen x^2 ) = (2x cos x² )/(sen x^2 ) = 2x cot⁡x²

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 + cos⁡x² e^(x^2 ). 2x) =

= [4x³ e^4x+ 3x²e^4x ] - 2x e^(x^2 ) 〖 sen〗⁡x² + 2xe^(x^2 ) cos⁡x² =

=4x³ e^4x + 3x²e^4x - 2x e^(x^2 ) sen⁡x² + 2xe^(x^2 ) cos⁡x² =

= x² e^4x(4x + 3) - 2x e^(x^2 )(sen⁡x²- cos⁡x²)

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〗

senh⁡x cosh⁡y + cosh⁡x senh⁡y = [(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〗

cos⁡x = -sen⁡〖k π/2-x〗

tan⁡〖x+y=((sen x)/cosx + (sen y)/cos⁡y )/(1-(sen x)/cos⁡x * (sen y)/cos⁡y )〗 = ((sen x 〖 cos〗⁡〖y +sen y cos⁡x 〗)/cos⁡〖x cos⁡y 〗 )/((cosx cos⁡〖y-senx sen y〗)/(cosx cos⁡y )) = (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)〗 =

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