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Aproxima el valor de la derivada de la siguiente función cuando x=1 usando diferencias
progresivas, regresivas y centrales. Utiliza en todos los casos h=0.1.Calcula el error cometido
comparando con el valor exacto al sustituir en la derivada de la función que se te indica
Valor real de la derivada =
−18+50𝑥+18𝑥2
−18𝑥3
(𝑥2+3)3
−18 + 50(1)+ 18(1)2
− 18(1)3
((1)2 + 3)3
−18 + 50 + 18 − 18
(1 + 3)3
−18 + 50
(4)3
32
64
= 0.5
Diferencias progresivas :
f’(1)=( f(1+0.1) – f(1) ) / 0.1
f(1.1) = (
3𝑥−1
𝑥2 +3
)2
= (
3(1.1)−1
1.12+3
)2
= (
2.3
4.21
)2
= 𝟎.𝟐𝟗𝟖𝟒𝟔𝟑𝟔𝟕𝟑𝟕𝟓𝟒𝟗𝟒𝟑𝟖𝟕
f(1)= (
3(1)−1
12+3
)2
= (
2
4
)2
= 0.25
f’(1)=( 0.29846367375494387–0.25) / 0.1 = 0.4846367375494387
Diferencias regresivas
f’(1)=( f(1) - f(1- 0.1) ) / 0.1 = ( f(1) - f(0.9) ) / 0.1
f(0.9) = (
3𝑥−1
𝑥2 +3
)2
= (
3(0.9)−1
0.92 +3
)2
= (
1.7
3.81
)2
= 0.19908928706746304
f’(1)=( 0.25 - 0.19908928706746304)/ 0.1 = 0.5091071293253696
Diferencias centrales
f’(1)= ( f(1.1) – f(0.9) )/ 2h
= ( 0.29846367375494387 - 0.19908928706746304 ) / 2*0.1
= 0.09937438668748083/0.2 = 0.49687193343740416
Valor real de la derivada = 0 .5 Error
Progresivas 0.4846367375494387 0.0153632624505613
Regresivas 0.5091071293253696 0.009107129325369612
Centrales 0.49687193343740416 0.0031280665625958437

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133467 p3a1

  • 1. Aproxima el valor de la derivada de la siguiente función cuando x=1 usando diferencias progresivas, regresivas y centrales. Utiliza en todos los casos h=0.1.Calcula el error cometido comparando con el valor exacto al sustituir en la derivada de la función que se te indica Valor real de la derivada = −18+50𝑥+18𝑥2 −18𝑥3 (𝑥2+3)3 −18 + 50(1)+ 18(1)2 − 18(1)3 ((1)2 + 3)3 −18 + 50 + 18 − 18 (1 + 3)3 −18 + 50 (4)3 32 64 = 0.5
  • 2. Diferencias progresivas : f’(1)=( f(1+0.1) – f(1) ) / 0.1 f(1.1) = ( 3𝑥−1 𝑥2 +3 )2 = ( 3(1.1)−1 1.12+3 )2 = ( 2.3 4.21 )2 = 𝟎.𝟐𝟗𝟖𝟒𝟔𝟑𝟔𝟕𝟑𝟕𝟓𝟒𝟗𝟒𝟑𝟖𝟕 f(1)= ( 3(1)−1 12+3 )2 = ( 2 4 )2 = 0.25 f’(1)=( 0.29846367375494387–0.25) / 0.1 = 0.4846367375494387 Diferencias regresivas f’(1)=( f(1) - f(1- 0.1) ) / 0.1 = ( f(1) - f(0.9) ) / 0.1 f(0.9) = ( 3𝑥−1 𝑥2 +3 )2 = ( 3(0.9)−1 0.92 +3 )2 = ( 1.7 3.81 )2 = 0.19908928706746304 f’(1)=( 0.25 - 0.19908928706746304)/ 0.1 = 0.5091071293253696
  • 3. Diferencias centrales f’(1)= ( f(1.1) – f(0.9) )/ 2h = ( 0.29846367375494387 - 0.19908928706746304 ) / 2*0.1 = 0.09937438668748083/0.2 = 0.49687193343740416 Valor real de la derivada = 0 .5 Error Progresivas 0.4846367375494387 0.0153632624505613 Regresivas 0.5091071293253696 0.009107129325369612 Centrales 0.49687193343740416 0.0031280665625958437