求函數(shù)y=ln[tan(x/2)]-cosx/[3(sin³x)]的導(dǎo)數(shù)
y′=[tan(x/2)]′/tan(x/2)-(1/3)(-sin⁴x-3cos²xsin²x)/sin⁶x
=[sec²(x/2)]/[2tan(x/2)]+(sin⁴x+3cos²xsin²x)/(3sin⁶x)
=cscx+(3-2sin⁴x)/(3sin⁶x)
如果題目是y=ln[tan(x/2)]-cos(x/3)sin³x,則：
y′=[tan(x/2)]′/tan(x/2)-[-(1/3)sin(x/3)sin³x+3cos(x/3)sin²xcosx]
=[sec²(x/2)]/[2tan(x/2)]+(1/3)sin(x/3)sin³x-3cos(x/3)sin²xcosx
=cscx+(1/3)sin(x/3)sin³x-3cos(x/3)sin²xcosx