10 急求几道大一高数题~要有步骤~感谢好心人~~~~ 实在是不会做~谢谢谢谢!
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sin'(x+y)=sin'x+sin'y
cos(x+y)*(1+y')=cosx+
cosy*y'
y'=[cosx-cos(x+y)]/[cos(x+y)-cosy]
xy+x=y-1
y=(x+1)/(1-x)
y'=[(1-x)-(x+1)*(-1)]/(1-x)^2
=2/(x-1)^2
5y^4*y'=siny+xcosy*y'+6y*y'
y'=siny/(xcosy+6y-5y^4)
y+xy'=e^(x+y)*(1+y')
y'=[y-e^(x+y)]/[e^(x+y)-x]
(x+y)^(x+1)*[ln(x+y)+(x+1)/(x+y) *(1+y')
=3+2y'
代入(0,2)
2*[ln2+1/2(1+y')]=3+2y'
y'=2ln2-2
cos(x+y)*(1+y')=cosx+
cosy*y'
y'=[cosx-cos(x+y)]/[cos(x+y)-cosy]
xy+x=y-1
y=(x+1)/(1-x)
y'=[(1-x)-(x+1)*(-1)]/(1-x)^2
=2/(x-1)^2
5y^4*y'=siny+xcosy*y'+6y*y'
y'=siny/(xcosy+6y-5y^4)
y+xy'=e^(x+y)*(1+y')
y'=[y-e^(x+y)]/[e^(x+y)-x]
(x+y)^(x+1)*[ln(x+y)+(x+1)/(x+y) *(1+y')
=3+2y'
代入(0,2)
2*[ln2+1/2(1+y')]=3+2y'
y'=2ln2-2
追问
谢谢!
追答
5y^4*y'=siny+xcosy*y'+6y*y'
y'=siny/(5y^4-xcosy-6y)
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