Question

Найти  $\frac{d^2y}{dx^2}$, где $y=y(t)=(t-1)e^t, x=x(t)=te^t$

Answer 1

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} =$
$= \frac{e^t + (t-1)e^t}{e^t + te^t} = \frac{te^t}{(1+t)e^t}=\frac{t}{t+1}$
$\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} =$
$= \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}$ = W.
$\frac{d}{dt} \frac{dy}{dx} = \frac{d}{dt}\frac{t}{t+1} =$
$= \frac{(t+1)-t}{(t+1)^2} = \frac{1}{(t+1)^2} = (t+1)^{-2}$
$\frac{dx}{dt} = e^t + te^t = (t+1)e^t$.
W= $= \frac{(t+1)^{-2}}{(t+1)e^t} = \frac{1}{(t+1)^3\cdot e^t}$