例1

已知{x=tety=2t+t2\left\{\begin{matrix}x=te^t\\y=2t+t^2\end{matrix}\right.{x=tety=2t+t2​，求dydx\dfrac{dy}{dx}dxdy​

解：
dxdt=(t+1)et\qquad\dfrac{dx}{dt}=(t+1)e^tdtdx​=(t+1)et

dydt=2+2t\qquad\dfrac{dy}{dt}=2+2tdtdy​=2+2t

dydx=dydtdxdt=2+2t(t+1)et=2et\qquad\dfrac{dy}{dx}=\dfrac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}=\dfrac{2+2t}{(t+1)e^t}=\dfrac{2}{e^t}dxdy​=dtdx​dtdy​​=(t+1)et2+2t​=et2​

例2

已知{x=3t2+2teysin⁡t−y+1=0\left\{\begin{matrix}x=3t^2+2t\\e^y\sin t-y+1=0\end{matrix}\right.{x=3t2+2teysint−y+1=0​，求dydx\dfrac{dy}{dx}dxdy​

解：
dxdt=6t+2\qquad\dfrac{dx}{dt}=6t+2dtdx​=6t+2

\qquad对eysin⁡t−y+1=0e^y\sin t-y+1=0eysint−y+1=0两边同时对ttt求导得

eyy′sin⁡t+eycos⁡t−y′=0\qquad e^yy'\sin t+e^y\cos t-y'=0eyy′sint+eycost−y′=0

\qquad移项得(eysin⁡t−1)y′=−eycos⁡t(e^y\sin t-1)y'=-e^y\cos t(eysint−1)y′=−eycost

dydt=y′=eycos⁡t1−eysin⁡t\qquad \dfrac{dy}{dt}=y'=\dfrac{e^y\cos t}{1-e^y\sin t}dtdy​=y′=1−eysinteycost​

dydx=dydtdxdt=eycos⁡t(6t+2)(1−eysin⁡t)\qquad\dfrac{dy}{dx}=\dfrac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}=\dfrac{e^y\cos t}{(6t+2)(1-e^y\sin t)}dxdy​=dtdx​dtdy​​=(6t+2)(1−eysint)eycost​

这题不仅考了参数方程求导，还考了隐函数求导。