求导：x=cos^4*t,y=sin^4*t,求dy/dx
\u8bbef(x)\u53ef\u5bfc\uff0cy=f(sin^2)+f(cos^2)\uff0c\u6c42\u5bfc\u6570dy/dxdy/dx=[d(ln cos e^x) / d(cos e^x)] \u00d7 [d(cos e^x) / e^x] \u00d7 [d(e^x) / x]
=[1/(cos e^x)] \u00d7 [- sin e^x] \u00d7 [e^x]
= - (tan e^x) \u00d7 e^x
dy/dt=4(sint)^3 *(sint)'
而
(cost)'= -sint
(sint)'=cost
于是
dy/dx
=(dy/dt) /(dx/dt)
=4(cost)^3 *(-sint) / [4(sint)^3 *cost]
= -(cot t)^2
母鸡模具及模具i
绛旓細dx/dt= 4(cost)^3 *(cost)'dy/dt=4(sint)^3 *(sint)'鑰(cost)'= -sint(sint)'=cost浜庢槸dy/dx=(dy/dt) /(dx/dt)=4(cost)^3 *(-sint) / [4(sint)^3 *cost]= -(cot t)^2
绛旓細dx/dt= 4(cost)^3 *(cost)'dy/dt=4(sint)^3 *(sint)'鑰(cost)'= -sint(sint)'=cost浜庢槸dy/dx=(dy/dt) /(dx/dt)=4(cost)^3 *(-sint) / [4(sint)^3 *cost]= -(cot t)^2
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