Question

найти производную dy/dx {x= ctg(2e^t)
{y=ln(tge^t)

Answer 1

$\left \{ {{x=ctg(2e^{t})} \atop {y=ln(tge^{t})}} \right. \\\\y'{x}=\frac{y'_{t}}{x'_{t}}=\frac{\frac{1}{tge^{t}}\cdot \frac{1}{cos^2e^{t}}\cdot e^{t}}{-\frac{1}{sin^2(2e^{t})}\cdot 2e^{t}}=-\frac{sin^2(2e^{t})}{2sine^{t}\cdot cose^{t}}=-\frac{sin^2(2e^{t})}{sin(2e^{t})}=-sin(2e^{t})$