Question

найти dy/dx
$\left \{ {{x=lny} \atop {y=} \frac{1}{} t^{2}-5 } \right.$

Answer 1

$\left \{ {{x=ln t} \atop {y=\frac{1}{t^2-5}}} \right. \\\\x'_{t}=\frac{1}{t}\\\\y'_{t}=- \frac{2t}{(t^2-5)^2} \\\\y'_{x}= \frac{dy}{dx}= \frac{y'_{t}}{x'_{t}} =- \frac{2t^2}{(t^2-5)^2}$