Matemática II EAP Ciencias Contables y Financieras

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(2x – 1)1/2 h’(x) -------------- (4)[pic 62][pic 63][pic 64][pic 65]

(x-2)3

Derivación de la ecuación de (2)

(x – 2)3 = 3(x – 2) = 3(x – 2)2[pic 66][pic 67][pic 68]

(2x – 1 )1/2 = ½ (2x – 1)-1/2 . 2 = (2x – 1)-1/2 -------------- (5)[pic 69][pic 70][pic 71][pic 72][pic 73]

De (4) (2x – 1)1/2 h’ (x) ------------ (6) [pic 74][pic 75]

(x – 2)3

Aplicando propiedad (1) reemplazando las ecuaciones derivadas (5)

h (x) = (x – 2)3 [pic 76]

(2x – 1)1/2

h’(x) = 3(x – 2)2 (2x – 1)1/2 - (x – 2)3 (2x – 1)-1/2 -------- (7)[pic 77]

(2x – 1)

h’(x) = (x – 2)2 3 (2x – 1)1/2 – (x – 2) ( 2x – 1)1/2[pic 78][pic 79][pic 80]

(2x – 1)

h’(x) = (x – 2)2 3(2x – 1) – (x – 2) = (x – 2)2 ( 5x - 1) ---- (8)[pic 81][pic 82][pic 83][pic 84][pic 85][pic 86]

(2x – 1) (2x – 1)1/2 (2x – 1) (2x – 1)1/2

Reemplazando (8) en (6)

(2x – 1)1/2 = (x – 2)2 (5x – 1) = (5x – 1)[pic 87][pic 88][pic 89][pic 90][pic 91][pic 92][pic 93]

(x – 2)3 (2x – 1) (2x – 1)1/2 (x – 2) (2x – 1)

[pic 94][pic 95][pic 96][pic 97][pic 98][pic 99][pic 100]

In (x – 2)3 = (5x – 1)[pic 101][pic 102][pic 103]

2x – 1 (x – 2) (2x – 1)[pic 104][pic 105]

05. Derivar implícitamente:

[pic 106]

d (x2 x) – d (xy2) + d (y2 ) = 0[pic 107][pic 108][pic 109]

dx dx dx

x2 dy + yd2 - xdy2 + y2dx + dy2 = 0[pic 110][pic 111][pic 112][pic 113][pic 114][pic 115][pic 116]

dx dx dx dx dx

x2dx + y2x - x 2ydy + y2 + 2ydy = 0[pic 117][pic 118][pic 119][pic 120][pic 121]

dx dx dx

x2dy + 2xy - 2xydy - y2 + 2xdy = 0[pic 122][pic 123][pic 124][pic 125][pic 126][pic 127][pic 128]

dx dx dx[pic 129][pic 130][pic 131]

x2 dy - 2xydy + 2xdy = y2 - 2xy

dx dx dx[pic 132][pic 133][pic 134]

dy ( x2 - 2xy + 2y) = y2 - 2xy[pic 135]

dx

dx = y2 - 2xy

dx x2 - 2xy + 2y [pic 136][pic 137]