Question

ПОЖАЛУЙСТА!! мАТЕМАТИКА
1 и 2 задания

Answer 1

$1)\; lim_{x\to \infty}(\frac{1+2x}{2x})^{\frac{x}{4}}=lim_{x\to \infty}(1+\frac{1}{2x})^{2x\cdot \frac{x}{4\cdot 2x}}=\\\\=lim_{x\to \infty}((1+\frac{1}{2x})^{2x})^{\frac{1}{8}}=e^{\frac{1}{8}}\\\\2)\; x+y=tgy\\\\x'+y'=(tgy)'\\\\1+y'=\frac{1}{cos^2y}\cdot y'\\\\y'(1-\frac{1}{cos^2y})=-1\\\\y'=\frac{-1}{1-\frac{1}{cos^2y}}=\frac{-1\cdot cos^2y}{cos^2y-1}=\frac{cos^2y}{1-cos^2y}=\frac{cos^2y}{sin^2y}=ctg^2y$