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

解： d x d t = ( t + 1 ) e t \qquad\dfrac{dx}{dt}=(t+1)e^t dtdx​=(t+1)et

d y d t = 2 + 2 t \qquad\dfrac{dy}{dt}=2+2t dtdy​=2+2t

d y d x = d y d t d x d t = 2 + 2 t ( t + 1 ) e t = 2 e t \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​

已知 { x = 3 t 2 + 2 t e y sin ⁡ 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 d y d x \dfrac{dy}{dx} dxdy​

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

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

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

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

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

d y d x = d y d t d x d t = e y cos ⁡ t ( 6 t + 2 ) ( 1 − e y sin ⁡ 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。