设f '(sinx^2)=cos2x+tanx^2 (0
问题描述：
设f '(sinx^2)=cos2x+tanx^2 (0
答
f '(sinx^2)=cos2x+tanx^2=1-2sinx^2+(sinx^2)/(1-sinx^2)
therefore
f'(x)=1-2x+x/(1-x)=1/(1-x)-2x
f(x)=-ln(1-x)-x^2