当我们知道 d x d t , d y d t \dfrac{dx}{dt},\dfrac{dy}{dt} dtdx,dtdy时，我们可以求出 d y d x \dfrac{dy}{dx} dxdy
d y d x = d y d t d x d t \dfrac{dy}{dx}=\dfrac{\quad \frac{dy}{dt}\quad}{\frac{dx}{dt}} dxdy=dtdxdtdy
然后我们也可以求 d 2 y d x 2 \dfrac{d^2y}{dx^2} dx2d2y
d 2 y d x 2 = d ( d y d x ) d t / d x d t \dfrac{d^2y}{dx^2}=\dfrac{d(\frac{dy}{dx})}{dt}/\dfrac{dx}{dt} dx2d2y=dtd(dxdy)/dtdx
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\qquad \dfrac{dx}{dt}=\dfrac{2t}{1+t^2}
dtdx=1+t22t
d y d t = 1 − 1 1 + t 2 = t 2 1 + t 2 \qquad\dfrac{dy}{dt}=1-\dfrac{1}{1+t^2}=\dfrac{t^2}{1+t^2} dtdy=1−1+t21=1+t2t2
d y d x = d y d t d x d t = t 2 \qquad \dfrac{dy}{dx}=\dfrac{\quad \frac{dy}{dt}\quad}{\frac{dx}{dt}}=\dfrac t2 dxdy=dtdxdtdy=2t
d ( d y d x ) d t = 1 2 \qquad \dfrac{d(\frac{dy}{dx})}{dt}=\dfrac 12 dtd(dxdy)=21
d 2 y d x 2 = d ( d y d x ) d t / d x d t = 1 2 2 t 1 + t 2 = 1 + t 2 4 t \qquad \dfrac{d^2y}{dx^2}=\dfrac{d(\frac{dy}{dx})}{dt}/\dfrac{dx}{dt}=\dfrac{\frac 12}{\frac{2t}{1+t^2}}=\dfrac{1+t^2}{4t} dx2d2y=dtd(dxdy)/dtdx=1+t22t21=4t1+t2