z=tan(3t+2x^2-y) x=1/t y=根号t 求dz/dt
问题描述：
z=tan(3t+2x^2-y) x=1/t y=根号t 求dz/dt
答
x=1/t,y=√t,u=3t+2x^2-yz=tan(3t+2x^2-y)=tanudz/du=(secu)^2du/dt=3,du/dx=4x,du/dy=-1dx/dt=-1/t^2,dy/dt=1/(2√t)dz/dt=dz/du*(du/dt+du/dx*dx/dt+du/dy*dy/dt)=(secu)^2*[3+4x*(-1/t^2)-1*1/(2√t)]=[sec(3t+2x...