z=tan(3t+2x^2-y) x=1/t y=根号t 求dz/dt
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z=tan(3t+2x^2-y) x=1/t y=根号t 求dz/dt
x=1/t,y=√t,u=3t+2x^2-y
z=tan(3t+2x^2-y)=tanu
dz/du=(secu)^2
du/dt=3,du/dx=4x,du/dy=-1
dx/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^2-y)]^2*[3-4/t^3-1/(2√t)]
z=tan(3t+2x^2-y)=tanu
dz/du=(secu)^2
du/dt=3,du/dx=4x,du/dy=-1
dx/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^2-y)]^2*[3-4/t^3-1/(2√t)]
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