f'(sin x)=1-cos x 求f''(x)
问题描述:
f'(sin x)=1-cos x 求f''(x)
答
f'(sinx)=1-根号(1-sin^2x)
f'(x)=1-根号(1-x^2)
f"(x)=-(-2x)/(2根号(1-x^2))
=x/根号(1-x^2)