Question

已知y=ln(x+√x∧2+1)arcsinx,求y'

Answer 1

arcsinx的导数是1/√(1-x^2)

所以

y'= [ln(x+√1+x^2)]' *arcsinx + ln(x+√1+x^2) * (arcsinx)'

= 1/(x+ √1+x^2) * (1+ x/√1+x^2)*arcsinx + ln(x+√1+x^2) * 1/√(1-x^2)

= 1/√(1+x^2) *arcsinx + ln(x+√1+x^2) * 1/√(1-x^2)