Question

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

Answer 1

$\left \{ {{x=te^{t}\quad } \atop {y=(t-1)e^{t}}} \right. \\\\x'_{t}=e^{t}+te^{t}=e^{t}(1+t)\\\\y'_{t}=e^{t}+(t-1)e^{t}=e^{t}(1+t-1)=te^{t}\\\\ \frac{dy}{dx} = \frac{y'_{t}}{x'_{t}}=\frac{te^{t}}{e^{t}(1+t)} =\frac{t}{t+1} \\\\y''_{tt}=e^{t}+te^{t}=e^{t}(1+t)\\\\ \frac{d^2y}{dx^2}= \frac{y''_{tt}}{x'_{t}} = \frac{e^{t}(1+t)}{e^{t}(1+t)} =1$