y=xarcsin(x/2)+根号(4-x平方),求导数,麻烦写详细点,

y=xarcsin√[x/(1+x)]+arctan√(x-√2)-√x,求导
dy/dx=arcsin√[x/(1+x)]+x{√[x/(1+x)]}′/√[1-x/(1+x)]+[√(x-√2)]′/(1+x-√2)-1/(2√x)
=arcsin√[x/(1+x)]+{x[x/(1+x)]′/2√[x/(1+x)]}/√[1/(1+x)]+{1/[2√(x-√2)]}/(x+1-√2)-1/(2√x)
=arcsin√[x/(1+x)]+{x[1/(1+x)²]/2√[x/(1+x)]}/√[1/(1+x)]+1/[2(x+1-√2)√(x-√2)]-1/(2√x)
=arcsin√[x/(1+x)]+(√x)/[2(1+x)]+1/[2(x+1-√2)√(x-√2)]-1/(2√x)

y'=arcsin(x/2)+{x/√[1-(x²/4)]}*(1/2)+1/2√(4-x²)*(-2x)
=arcsin(x/2)

先整理一下,设x/2=sint 那么,y=+2cost 然后 y'=(2sint*t)'+(2cost)'
(2cost)'=-2sint (2sint*t)'=2(sint')*t+2sint*t'=2cost*t+2sint
所以y'=-2sint+2cost*t+2sint=2cost

y=xarcsin(x/2)+√(4-x^2),
y'=[xarcsin(x/2)]'+[√(4-x^2)]',
=arcsin(x/2)+x*1/2*1/√(1-x^2/4)+1/2*(-2x)*1/√(4-x^2),
=arcsin(x/2)+x/√(4-x^2)-x/√(4-x^2),
=arcsin(x/2)
解毕