为自己加油

设z=x^y，而x=sint,y=cos，求dz/dt．

∂z/∂x=(x ^y)′_x，=yx^y-1，∂z/∂y=(x^y)′_y=x^ylnx dx/dt=(sint)′=cost，(dy/dt)=(cost)′=-sint 所以dz/dt=∂z/∂x•dx/dt+∂z/∂y•dy/dt =yx^y-1•cost+x^ylnx•(-sint) =cos²t•(sint)^cost-1-(sint)^cost+1•lnsint =(sint)^cost-1[cos²t-(sint)²•lnsint]