CD_U3_A4_FRPB

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NOMBRE: FRANCISCO JAVIER PRIEGO BRITO FACILITADOR: OSCAR RICARDO DELFIN SANTIESTEBAN MATERIA: CÁLCULO DIFERENCIAL CARRERA: DESARROLLO DE SOFTWARE TEMA: DERIVACION DE FUNCIONES ALGEBRAICAS

Transcript of CD_U3_A4_FRPB

Page 1: CD_U3_A4_FRPB

NOMBRE: FRANCISCO JAVIER PRIEGO BRITO

FACILITADOR: OSCAR RICARDO DELFIN SANTIESTEBAN

MATERIA: CÁLCULO DIFERENCIAL

CARRERA: DESARROLLO DE SOFTWARE

TEMA: DERIVACION DE FUNCIONES ALGEBRAICAS

Page 2: CD_U3_A4_FRPB

f ' (x)= lim∆x→0

5 (x+∆ x)+3−5 x−3∆x

f ' (x)= lim∆x→0

5 x+5∆ x+3−5 x−3∆ x

f ' (x)= lim∆x→ 0

5 ∆x∆ x

f ' (x)= lim∆x→0

¿5; f ' (x)=5

f ' (x)= lim∆x→0

¿¿

f ' (x)=lim∆ x→0

x2+2 x ∆x+∆x2+1−x2−1

∆ x

f ' (x)=lim∆ x→0

2 x ∆ x+∆ x2

∆ x

f ' (x)= lim∆x→0

¿)= lim∆ x→ 0

(2 x ∆ x∆ x

)+ lim∆ x→0¿)

f ' (x)= lim∆x→0

2x+ lim∆ x→0

∆ x

f ' (x)=2x

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f ' (x)= lim∆x→0

5 x+5∆ x+104−5 x−104∆ x

f ' (x)= lim∆x→ 0

5 ∆x∆ x

f ' ( x )=5

f ' ( x )= lim∆ x→0

¿¿

f ' (x)=lim∆ x→0

x2+2 x ∆x+∆x2−1−x2+1

∆ x

f ' (x)=lim∆ x→0

2 x ∆ x+∆ x2

∆ x

f ' (x)= lim∆x→0

(2 x+∆ x )

f ' (x)=2x

f ' (x)= lim∆x→0

20 ( x+∆ x )+3 /4−20 x−3/4∆ x

f ' (x)= lim∆x→ 0

20 x+20∆ x+3/ 4−20x−3/4∆ x

f ' (x)= lim∆x→0

20 ∆ x∆ x

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f ' (x)=20

f ' (x)= lim∆x→0

8¿¿

f ' (x)=lim∆ x→0

8x2+16 x ∆ x+8∆ x2+2−8 x2−2

∆ x

f ' (x)=lim∆ x→0

16 x ∆ x+8∆ x2

∆ x

f ' ( x )= lim∆ x→0

16 x+ lim∆ x→0

8 ∆ x

f ' ( x )=16 x

f ' (x)= lim∆x→0

35 ( x+∆ x )

−9− 35 x

+9

∆ x

f ' (x)= lim∆x→0

3 x5

+ 3∆ x5

−9−3 x5

+9

∆ x

f ' (x)= lim∆x→0

3 ∆ x5 ∆ x

f ' (x)=3/5

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f ' (x)= lim∆x→0

5/2¿¿

f ' (x)=lim∆ x→0

52x2+5 x ∆x+∆ x2−100−5

2x2+100

∆ x

f ' (x)=lim∆ x→0

5 x ∆ x+∆ x2

∆ x

f ' ( x )= lim∆ x→0

(5 x+∆ x )

f ' ( x )=(5 x )

f ' (x)= lim∆x→0

910 ( x+∆ x )

+9− 910x

−9

∆ x

f ' (x)= lim∆x→0

9 x10

+ 9∆ x10

+9−9 x10

−9

∆ x

f ' (x)= lim∆x→0

9 x10

+ 9∆ x10

+9−9 x10

−9

∆ x

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f ' (x)= lim∆x→0

9∆ x10 ∆x

f ' (x)=9/10

f' ( x )= lim

∆ x→0

3( x+∆ x )

−3x

∆x

f ' ( x )= lim∆ x→0

3x−3 ( x+∆x )x ( x+∆ x )∆ x

f ' ( x )= lim∆ x→0

−3∆ xx ( x+∆x )∆ x

f ' ( x )= lim∆ x→0

−3 ∆xx2∆x+x ∆ x2

¿¿

f ' (x)= lim∆x→ 0

[ x2∆ x+x ∆ x2

−3∆ x]−1

f ' (x)= lim∆x→ 0

[ x2+x ∆ x−3

]−1

f ' (x)= lim∆x→ 0

( −3x2+ x∆ x

)

f ' (x)=lim∆ x→0

−3

x2

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f ' ( x )=−3 x−2

f ' (x)= lim∆x→0

53¿¿

¿¿

f ' (x)= lim∆x→ 0

5 x2−5¿¿¿¿

f ' (x)= lim∆x→0

5 x2−5 x2−10 x ∆x−5∆ x2

3x2 ¿¿¿

f ' (x)= lim∆x→ 0

5 x2−5 x2−10 x ∆x−5∆ x2

3x2 ¿¿¿

f ' (x)= lim∆x→0

−10 x ∆ x−5∆ x2

3 x2¿¿¿

f ' (x)= lim∆x→ 0

−10 x−5 ∆ x3 x2¿¿

¿

f ' ( x )=−10 x /3 x4=

f ' ( x )=−103 x3

f ' (x)= lim∆x→0

2¿¿

f ' (x)=lim∆ x→0

2 x3+6 x2∆ x+6 x ∆ x2+∆ x3+5−2 x3−5

∆ x

f ' (x)=lim∆ x→0

6x2∆ x+6 x ∆ x2+∆x3

∆x

f ' (x)= lim∆x→0

6 x2

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f ' (x)=6 x2

f ' (x)= lim∆x→0

a(x+∆ x )+N−ax−N∆ x

f ' (x)= lim∆x→ 0

ax+a∆ x+N−ax−N∆x

f ' (x)= lim∆x→0

a∆ x∆ x

f ' (x)=a

f ' (x)= lim∆x→0

a¿¿

f ' (x)=lim∆ x→0

a x2+2a x ∆ x+a∆ x2+N−ax2−N

∆ x

f ' (x)=lim∆ x→0

2a x ∆ x+a∆ x2

∆ x

f ' ( x )= lim∆ x→0

( 2ax∆ x

+a∆ x )

f ' ( x )=2ax

f ' (x)= lim∆x→0

a¿¿

f ' (x)=lim∆ x→0

ax3+2ax2∆ x+2ax ∆ x2+a∆ x3+N−ax3−N

∆ x

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f ' ( x )=lim∆ x→0

2ax2∆ x+2ax ∆ x2+a∆ x3

∆ x

f ' ( x )=2a x2

¿

f ' (x)= lim∆x→0

a¿¿

Donde (nk) son constantes

f ' (x)=lim∆ x→0

a xn+(n1)∆ x .a xn−1+(n2)∆x2 . a xn−2+…+( nn−1)ax∆ xn−1+(nn)a ∆ xn−a xn∆ x

f ' (x)=lim∆ x→0 (n1)∆ x .a xn−1+(n2)∆x2 . a xn−2+…+( nn−1)ax∆ xn−1+(nn)a∆ xn

∆ x

f ' (x)=lim∆ x→0 (n1)∆ x .a xn−1

∆ x+(n2)∆ x2 . a xn−2

∆ x+…+( nn−1)ax∆ xn−1

∆ x+(nn)a∆ xn

∆ x

f ' (x)= lim∆x→0

na xn−1+(n2)∆ x .axn−2+…+( nn−1)ax∆ xn−2+(nn)a ∆ xn−1

f ' ( x )=an xn−1

p) a

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Si a es una constante entonces significa que no existe derivada.

Que: dxn /dx= nxn-1

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Ejercicio 2. Calcula la derivada de las funciones.

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Ejercicio 3. Calcula la derivada de las funciones.

u=x3 ; v=(x-1); u’=3x2; v’=1Sustituyendo nos da:

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f(x)=(x+1); f’(x)=1; g1(x)=(x-3); g1’(x)=1

g’(x)=(x-3)+(x+3); g’(x)=2x-2