Question

Вычислить производную
{y = In(arctgt), x= 1/(1+t^2)​

Answer 1

$\{ {{y=ln(arctg(t))} \atop{x=\frac{1}{1+t^2} }} \right. \\\{ {{\frac{dy}{dt} =\frac{1}{arctg(t)(1+t^2)} \atop{\frac{dx}{dt} =\frac{-2t}{(1+t^2)^2} }} \right.\\\frac{dy}{dx} =- \frac{1+t^2}{2t*arctg(t)}$