Question

Найти d^2y/dх^2 функции, заданной параметрически: Cистема: y=arcctgt y=ln(2+t^2)

Answer 1

$\left \{ {{y=arcctgt} \atop {x=ln(2+t^2)}} \right. \; \; \; \; \; \; \; y'_{x}= \frac{y'_{t}}{x'_{t}} \\\\y'_{t}=-\frac{1}{1+t^2}\\\\x'_{t}=\frac{2t}{2+t^2}\\\\y'_{x}=-\frac{2+t^2}{(1+t^2)\cdot 2t}\\\\y''_{xx}= \frac{(y'_{x})'_{t}}{x'_{t}} \\\\(y'_{x})'_{t}=(-\frac{2+t^2}{2t^3+2t})'_{t}=- \frac{2t(2t^3+2t)-(2+t^2)(6t^2+2)}{(2t^3+2t)^2} =\\\\=- \frac{4t^4+4t^2-12t^2-4-6t^4-2t^2}{4t^2\cdot (1+t^2)^2}= -\frac{-2t^4-10t^2-4}{4t^2(1+t^2)^2} = \frac{t^4+5t^2+2}{t^2(1+t^2)^2}$

$y''_{xx}= \frac{t^4+5t^2+2}{t^2(1+t^2)^2} : \frac{2t}{2+t^2} = \frac{(2+t^2)(t^4+5t^2+2)}{2t^3(1+t^2)^2}$