y=tan (x+y)求导
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2015-11-06
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y=tan(x+y)得y'=(1+y')/(cos(x+y))^2
解得y'=-1/(sin(x+y))^2=-(sin(x+y))^(-2)
y''=2(sin(x+y))^(-3)*cos(x+y)*(1+y')
=2(sin(x+y))^(-3)*cos(x+y)*(1-1/(sin(x+y))^2)
=-2(cos(x+y))^3(sin(x+y))^(-5)
解得y'=-1/(sin(x+y))^2=-(sin(x+y))^(-2)
y''=2(sin(x+y))^(-3)*cos(x+y)*(1+y')
=2(sin(x+y))^(-3)*cos(x+y)*(1-1/(sin(x+y))^2)
=-2(cos(x+y))^3(sin(x+y))^(-5)
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