dy/dx= [7x-2]^-2
y=?
2007-10-03 03:13:14 · 5 answers · asked by Chocolate Strawberries. 4 in Science & Mathematics ➔ Mathematics
dy/dx= 1/(7x -2 )^2
y
= ∫1/(7x -2 )^2 dx
= (1/7) ∫1/(7x -2 )^2 d(7x-2)
= -(1/7)/(7x-2) + c
2007-10-03 03:24:38 · answer #1 · answered by sahsjing 7 · 1⤊ 0⤋
â«(7x - 2)^-2 dx
= â«[1 / (7x - 2)^2] dx
Let y = (7x - 2)^-1
y' = -7(7x - 2)^-2
Adjust to suit the original question:
Let y = -1/7 (7x - 2) ^-1
y' = (7x - 2) ^ -2
Therefore:
y' or dy/dx = -1/7 (7x-2) ^-1 + c
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Hope it helps :)
2007-10-03 10:42:04 · answer #2 · answered by Micoaw 3 · 0⤊ 1⤋
I = â« (7x - 2)^(- 2 ) dx
Let u = 7x - 2
du / dx = 7
dx = du / 7
I = ( 1 / 7 ) â« u ^ (- 2) du
I = ( - 1 )( 1 / 7 ) (1 / u) + C
I = ( - 1 ) / [ 7 (7x - 2) ] + C
2007-10-03 10:54:22 · answer #3 · answered by Como 7 · 2⤊ 1⤋
dy/dx=[7x-2]^-2
y=integ [7x-2]^-2
7x-2=t
7dx=dt
dx=dt/7
(1/7)integ t^(-2)
(1/7)(t^(-1))/-1
=-1/(7t)
=-1/7[7x-2]+C
2007-10-03 10:37:17 · answer #4 · answered by cidyah 7 · 0⤊ 1⤋
dy/dx=(7x-2)^2
=49x^2-28x+4
y=(49x^3)/3-14x^2+4x
2007-10-03 10:27:24 · answer #5 · answered by Anonymous · 0⤊ 1⤋