Question

Найти y' и y'"
Найти y' и y"​

Answer 1

$\left\{\begin{array}{l}\ \, x=\dfrac{lnt}{t}\\y=t^2\, lnt\end{array}\right\\\\\\y'_{t}=2t\cdot lnt+t^2\cdot \dfrac{1}{t}=2t\cdot lnt+t=t\cdot (2\, lnt+1)\\\\x'_{t}=\dfrac{\frac{1}{t}\cdot t-lnt\cdot 1}{t^2}=\dfrac{1-lnt}{t^2}\\\\\\y'_{x}=\dfrac{y'_{t}}{x'_{t}}=\dfrac{t\cdot (2\, lnt+1)}{\dfrac{1-lnt}{t^2}}=\dfrac{t^3\cdot (2\, lnt+1)}{1-lnt}$

$(y'_{x})'_{t}=\dfrac{(6t^2\, lnt+2t^2+3t^2)(1-lnt)-(2t^3\cdot lnt+t^3)\cdot (-\frac{1}{t})}{(1-lnt)^2}=\\\\\\=\dfrac{(6t^2\, lnt+5t^2)(1-lnt)+2t^2\, lnt+t^2}{(1-lnt)^2}\\\\\\y''_{xx}=\dfrac{(y'_{x})'_{t}}{x'_{t}}=\dfrac{\dfrac{(6t^2\, lnt+5t^2)(1-lnt)+2t^2\, lnt+t^2}{(1-lnt)^2}}{\dfrac{1-lnt}{t^2}}=\\\\\\=\dfrac{(6t^4\, lnt+5t^4)(1-lnt)+2t^4\, lnt+t^4}{(1-lnt)^3}=\\\\\\=\dfrac{3t^4\, lnt+6t^4-6t^4\, ln^2t}{(1-lnt)^3}=\dfrac{3t^4\, (lnt-2ln^2t+2)}{(1-lnt)^3}$