CD_U3_A7_SABR
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Transcript of CD_U3_A7_SABR
1. Calcula la derivada de las siguientes funciones con respecto a x’.
a) x² + y² + 2xy = x
2x + 2y.y’+ 2y + 2xy’ = 1
2y.y’ +2xy.y’ = -2x-2y + 1
y’ ( 2y + 2xy) = -2x - 2y + 1
y’ = =
b) y sen x + = 3y
1.y’ cos x + 1.y’x + 1y = 3.y’
1.y’ + 1.y’x – 3.y’ = -cos x + 1y
y’ (1 + 1x – 3 ) = -cos x +1y
y’=
c) y tan x + x cos y = 1
y’ tanx +y sec² x + cos y - x sen y .y’= 0
Y’(tan x – xsen y) = y sec² x + cos y
y’ =
d) =
Xy(x) = 1( x² - y² )
x²+xy = x² - y²
2x + y + xy’ = 2x – 2y.y’
Xy’ + 2y.y’ = 2x -2x + y
y’(x + 2y) = y
y’=
e) + x = 2x²y²
x + y + x = 2x ² y² (x-y)x + y + x = 2x³y² - 2x²y³1 + y’ + 1 = 6x²y² + 2x³2y.y’ – 4xy³ - 2x²2y².y’y’ + 2 = 6x²y² + 2x³2y.y’ – 4xy³ - 2x²2y².y’y’ -2x³2y.y’ +2x²2y².y’ = 6x²y² - 4xy³ - 2y’ (1 – 2x³2y + 2x²2y²) = 6x²y² - 4xy³ - 2
y’= = = - =
y’ =- = - = - = -
f) Xysen xNo se puede derivar.
g) = 3y cos y
sen x = 3ycos y (y)sen x = 3y² cos ycos x = 6y.y’ cos y - 3y² sen y .y’-6y.y’ + 3y² sen y .y’ = cos xy’ ( 6y + 3y² sen y) = cos x
y’ =
h) = ln x y
y’ cosx + senx - (y sen x)’ =
y’ cos x + sen x –y’ senx – y cos x =
xy (y’ cos x –y’ senx ) = y+xy’ - sen x + ycos x
xy.y’ xy xycosx – xy.y’ xy senx = y - sen x + ycos x
y’ ( 3xy cos x – 2xy sen x - x²y) = y - sen x + ycos x
y’ =
y’ =