Question

y=COS(2X^2+1) y=ln[X+(1+X^2)^1/2] 的导数求解步走

Answer 1

y'=[2x^2+1]'[-sin(2x^2+1)]=-4x sin(2x^2+1)

y'=[x+(1+x^2)^1/2]'/[x+(1+x^2)^1/2]

=[1+1/2*(1+x^2)^(-1/2)*(1+x^2)']/[x+(1+x^2)^1/2]

=[1+x/(1+x^2)^(1/2)]/[x+(1+x^2)^1/2]