Question

求y=arcsin根号（1-x^2）的微分

Answer 1

y=arcsin√(1-x^2)
dy=[arcsin√(1-x^2)]'dx
=1/√[1-(√(1-x^2))^2]*(-x)/√(1-x^2)dx
=1/x*(-x)/√(1-x^2)dx
=-1/√(1-x^2)dx