Question

数学求导计算

求导y＝ln（x+根号下1减x平方）

Answer 1

y=ln[x+√(1-x^2)]
y'={1/[x+√(1-x^2)]}*[x+√(1-x^2)]'
={1/[x+√(1-x^2)]}*[1+(1/2)*(1-x^2)^(-1/2)*(1-x^2)']
={1/[x+√(1-x^2)]}*[1+(1/2)*(1-x^2)^(-1/2)*(-2x)]
={1/[x+√(1-x^2)]}*[1-x/√(1-x^2)]
={1/[x+√(1-x^2)]}*[√(1-x^2)-x]/√(1-x^2)
=[√(1-x^2)-x]/{[x+√(1-x^2)]*√(1-x^2)}