y=sin(x/2),求y^(n)
问题描述:
y=sin(x/2),求y^(n)
我做出来是(sin(x/2+nπ/2))/(2^n)答案是(sin(x+nπ/2))/(2^n)我带了个n=1进去,是要x/2的呀
答
因为y=sin(x)时
y^(n)=sin(x+nπ/2)
所以
y=sin(x/2),y^(n)=sin(x/2+nπ/2)/(2^n)当n=1时y'=1/2cos(x/2)y^(1)=sin(x/2+π/2)/(2^1)=1/2 cos(x/2)