DERIVACION DE ORDEN SUPERIO E IMPLICITAS

Guadalajara, Jal a 5 de Septiembre del 2014

Grupo: ER-ECDN-1402S-BI-001

Profesora: Eréndira Santos Viveros.

Alumno: Francisco Solís Mancilla

Materia: Calculo diferencial Unidad 3, Números reales y funciones

Actividad 3. Derivación de orden superior e implícita

Instrucciones: Determina la derivada de las funciones implícitas y de orden superior, además de realizar las demostraciones de las funciones presentadas.

Calcula las siguientes derivadas de funciones implícitas, suponiendo que depende de

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(d )/dx (sen^2 (xy)+xy^2 ) = d/x(x³+ 1)

( 2sen⁡xy *cos⁡〖xy*((x d(y))/dx〗 +y dx/dx )+ x d/dx(〖y)〗^2 + y² (d x)/dx = d/x(x³+ 1)

(2sen⁡xy * cos⁡〖xy*(x dy/dx〗 + y dx/dx)+ 2xy dy/dx + y² (d x)/dx = 3x²

(2sen⁡xy * cos⁡〖xy*(x dy/dx〗 +y dx/dx) +2xy dy/dx + y² = 3x²

(2sen⁡xy * cos⁡〖xy*(x dy/dx〗 + y) + 2xy dy/dx = 3x² - y²

2x sen⁡xy * cos⁡〖xy*dy/dx〗 + 2y sen⁡xy * cos⁡xy + 2xy dy/dx = 3x² - y²

dy/dx (2x sen⁡xy * cos⁡xy + 2xy) = 3x² - y² - 2y sen⁡xy * cos⁡xy

dy/dx = (3x² - y^2- 2y sen⁡xy * cos⁡xy)/(2x sen⁡xy * cos xy + 2xy)

Si deseamos simplificar se tiene el seno (2a) = 2 sin⁡a .cos⁡a

dy/dx = (3x² - y^2- y sen⁡2xy )/(x 〖sen 2〗⁡xy + 2xy)

.

Derivar si y depende de x realizar √(x+x²y ) + e^xy = ln⁡〖(x+y)〗

d/dx (〖x+x²y)〗^(1/2) + d/dx 〖(e〗^xy)= 〖 d/dx ln〗⁡〖(x+y〗)

1/2 (〖x+x²y)〗^((-1)/2) * d/dx (x+x²y) + e^xy (d xy)/dx = 1/(x+y) d/dx (x+y)

1/2 ((〖x+x²y)〗^((-1)/2) *( d/dx(x) + d/dx (x²y))) + e^xy (d xy)/dx = 1/(x+y) (d/dx (x) + d/dx (y))

1/2* 1/√(x+x²y) (1+ 2xy’) +( e^xy * x d/dx(y) + y d/dx(x)) = 1/(x+y) (1 + dy/dx)

1/(2√(x+x²y)) + 2xy'/(2√(x+x²y)) + e^xy(x y’ + y) = 1/(x+y) (1 +y’)

2xy'/(2√(x+x²y)) + e^xy(X y’ + y) = 1/(x+y) (1 +y’) - 1/(2√(x+x²y))

Y’ 2x/(2√(x+x²y)) + y’ x e^xy + ye^xy - 1/(x+y) – y’ 1/(x+y) = - 1/(2√(x+x²y))

Y’ 2x/(2√(x+x²y)) + y’ x e^xy - y’ 1/(x+y)= - 1/(2√(x+x²y)) - ye^xy + 1/(x+y)

Y’ (x/√(x+x²y) + x e^xy- 1/(x+y)) = - 1/(2√(x+x²y)) - ye^xy + 1/(x+y)

Y’ = (- 1/(2√(x+x²y)) - ye^xy + 1/(x+y))/((x/√(x+x²y) + x e^xy- 1/(x+y)) )

.

d/dx ( ln⁡(x/y) + d/dx (sen^2 (x+y²) = d/dx(2x2 y)

d/dx(ln⁡x - ln⁡y) + 2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗 d/dx (x+y²) = d/dx(2x2 y)

d/dx(ln⁡x - ln⁡y) + 2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗 d/dx (x+y²) = 2x² d/dx(y) + y d/dx(2x²)

1/x - 1/y * dy/dx +2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗 ( d/dx(x) + d/dx (y²) = 2x² dy/dx + 4xy

- 1/y * dy/dx +2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗(1 + 2y dy/dx) = 2x² dy/dx + 4xy - 1/x

- 1/y * dy/dx + 2 (1 + 2y dy/dx) (sen⁡(x+y²)* cos⁡〖(x+y²) 〗) = 2x² dy/dx + 4xy - 1/x

- 1/y * dy/dx + (2 + 4y dy/dx) (sen⁡(x+y²)* cos⁡〖(x+y²) 〗) = 2x² dy/dx + 4xy - 1/x

- 1/y * dy/dx + 2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗+ 4y sen⁡(x+y²)* cos⁡〖(x+y²) dy/dx 〗= 2x² dy/dx + 4xy - 1/x

- 1/y * dy/dx + 4y sen⁡(x+y²)* cos⁡〖(x+y²) dy/dx 〗- 2x² dy/dx = 4xy - 1/x - 2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗

dy/dx (- 1/y + 4y sen⁡(x+y²)* cos⁡〖 (x+y²)〗 - 2x²) = 4xy - 1/x - 2 sen⁡(x+y²)* cos⁡〖(x+y²) 〗

dy/dx (- 1/y + 4y sen⁡(x+y²)* cos⁡〖 (x+y²)〗 - 2x²) = 4x²y - 1 – 2x sen⁡(x+y²)* cos⁡〖(x+y²) 〗

(dy )/dx(-2x²y -1 + 4y² sen⁡(x+y²)* cos⁡〖 (x+y²)〗) = 4x²y - 1 – 2x sen⁡(x+y²)* cos⁡〖(x+y²) 〗

(dy )/dx= (4x²y - 1 – 2x sen⁡(x+y²)* cos⁡〖(x+y²) 〗)/(-2x²y -1 + 4y² sen⁡(x+y²)* cos⁡〖 (x+y²)〗 )

Calcular las siguientes derivadas de orden superior:

.

d/dx(x^5 e^3x) = x^(5 ) d/dx (e^3x) + e^3x d/dx (x^(5 ))

y´ (x^5 e^3x) = x^(5 ) e^3x * 3 + e^3x * 5〖 x〗^(4 )

y´ (x^5 e^3x) = 3〖 x〗^(5 ) e^3x + 5〖 x〗^(4 ) e^3x

y´ (x^5 e^3x) = e^3x (3〖 x〗^(5 )+ 5〖 x〗^4)

y´´ (x^5 e^3x) = e^3x d/dx(3〖 x〗^(5 )+ 5〖 x〗^4) + (3〖 x〗^(5 )+ 5〖 x〗^4) d/dx(e^3x)

y´´ (x^5 e^3x) = e^3x (15〖 x〗^(4 )+ 20〖 x〗^3) + (3〖 x〗^(5 )+ 5〖 x〗^4) e^3x * 3

y´´ (x^5 e^3x) = e^3x (15〖 x〗^(4 )+ 20〖 x〗^3) + 3 e^3x (3〖 x〗^(5 )+ 5〖 x〗^4)

y´´´ (x^5 e^3x) = e^3x d/dx (15〖 x〗^(4 )+ 20〖 x〗^3) + (15〖 x〗^(4 )+ 20〖 x〗^3) d/dx (e^3x) + 3 e^3x d/dx (3〖 x〗^(5 )+ 5〖 x〗^4) + (3〖 x〗^(5 )+ 5〖 x〗^4)d/dx(3 e^3x)

y´´´ (x^5 e^3x) = e^3x (60 〖 x〗^(3 )+ 60〖 x〗^2) + (15〖 x〗^(4 )+ 20〖 x〗^3) (e^3x)(3) + 3 e^3x (15〖 x〗^(4 )+ 20〖 x〗^3) + (3〖 x〗^(5 )+ 5〖 x〗^4) (3 * e^3x *3)

y´´´ (x^5 e^3x) = e^3x (60 〖 x〗^(3 )+ 60〖 x〗^2) + (e^3x) 3 (15〖 x〗^(4 )+ 20〖 x〗^3) + 3 e^3x (15〖 x〗^(4 )+ 20〖 x〗^3) + * e^3x 9 (3〖 x〗^(5 )+ 5〖 x〗^4)

y´´´ (x^5 e^3x) = e^3x (60 〖 x〗^(3 )+ 60〖 x〗^2) + (e^3x) (45〖 x〗^(4 )+ 60〖 x〗^3) + e^3x (45〖 x〗^(4 )+ 60〖 x〗^3) + e^3x (27〖 x〗^(5 )+ 45〖 x〗^4)

y´´´ (x^5 e^3x) = e^3x(60 〖 x〗^(3 )+ 60〖 x〗^2 + 45〖 x〗^(4 )+ 60〖 x〗^3+ 45〖 x〗^(4 )+ 60〖 x〗^3 + 27〖 x〗^(5 )+ 45〖 x〗^4)

y´´´ (x^5 e^3x) = e^3x (180 〖 x〗^(3 )+ 60 〖 x〗^2 + 135 〖 x〗^(4 ) + 27 〖 x〗^(5 ))

y´´´ (x^5 e^3x) = e^3x (27〖 x〗^(5 )+ 135 〖 x〗^(4 ) + 180 〖 x〗^(3 )+ 60 〖 x〗^2

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y´ (e^3x sen⁡2x) = e^3x d/dx (sen⁡2x) + sen⁡2x d/dx e^3x

y´ (e^3x sen⁡2x) = e^3x cos⁡2x * 2 + sen⁡2x * e^3x *3

y´ (e^3x sen⁡2x) = 2e^3x cos⁡2x + 3 〖e^3x sen〗⁡2x *

y´ (e^3x sen⁡2x) = e^3x (2cos⁡2x + 3 sen⁡2x)

y´´ (e^3x sen⁡2x) = e^3x d/dx (2〖 cos〗⁡2x + 3 sen⁡2x) + (2〖 cos〗⁡2x + 3 sen⁡2x) d/dx (e^3x ) =

y´´ (e^3x sen⁡2x) = e^3x ( d/dx (2〖 cos〗⁡2x) + d/dx (3 sen⁡2x)) + (2〖 cos〗⁡2x + 3 sen⁡2x) ( d/dx (e^3x )) =

y´´ (e^3x sen⁡2x) = e^3x ((- 2 sen⁡2x*2) + 3 〖 cos〗⁡2x *2) + (2〖 cos〗⁡2x + 3 sen⁡2x) (e^3x*3) =

y´´ (e^3x sen⁡2x) = e^3x(- 4 sen⁡〖 2x〗 + 6 〖 cos〗⁡2x ) + e^3x (6〖 cos〗⁡2x + 9 sen⁡2x) =

y´´ (e^3x sen⁡2x) = e^3x(- 4 sen⁡〖 2x〗 + 6 〖 cos〗⁡2x ) + e^3x (9 sen⁡2x

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