设函数y=sinx分之x加1,求y'?
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y'=(sin(1/x))'
=cos(x+1/x)*(x+1/x)'
=cos(x+1/x)*(1-1/x^2 ),2,解答如下: ,2,1/sin(x)+(x+1)*cos(x)/[sin(x)]^2,1,y'=sin^2 x分之sinx+cos(x+1),0,y=x/sinx+1
y'=(sin(1/x))'
=cos(1/x)*(1/x)'
=-cos(1/x)/x^2,0,
=cos(x+1/x)*(x+1/x)'
=cos(x+1/x)*(1-1/x^2 ),2,解答如下: ,2,1/sin(x)+(x+1)*cos(x)/[sin(x)]^2,1,y'=sin^2 x分之sinx+cos(x+1),0,y=x/sinx+1
y'=(sin(1/x))'
=cos(1/x)*(1/x)'
=-cos(1/x)/x^2,0,
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