还是求导..1)(arcctg1/x)^22)y=a根号x * x^23)y=x * arcsinx + 根号1-x^2
热心网友
1)y=[arcctg(1/x)]^2y'=[-1/(1+x^2)](1/x)' =[-1/(1+x^2)](-1/x^2) =1/(1+x^2)x^22)y=a√x * x^2=ax^(5/2)y'=a(5/2)x^(3/2)=(5a/2)x^(3/2)3)y=x * arcsinx + √(1-x^2)y'=arcsinx+[x/√(1-x^2)]+(1/2)(-2x)/√(1-x^2) =arcsinx