Question

Помогите пожалуйста.​

Answer 1

$1)\; \; \lim\limits _{x \to \infty}\frac{x^5+x^4+2x^2}{x^3-5x^5+x^4}=\lim\limits _{n \to \infty}\frac{1+\frac{1}{x}+\frac{2}{x^3}}{\frac{1}{x^2}-5+\frac{1}{x}}=-\frac{1}{5}\\\\2)\; \; \lim\limits _{x \to 20}\frac{\sqrt{x+5}-5}{x-2x+20}=\lim\limits _{x \to 20}\frac{x+5-25}{(-x+20)(\sqrt{x+5}+5)}=\lim\limits _{x \to 20}\frac{x-20}{-(x-20)(\sqrt{x+5}+5)}=\\\\=\lim\limits _{x \to 20}\frac{1}{-(\sqrt{x+5}+5)}=\frac{1}{-\sqrt{25}-5}=-\frac{1}{10}$

$3)\; \; y=sin^32x+cos^25x\\\\y'=3\cdot sin^22x\cdot cos2x\cdot 2+2cos5x\cdot (-sin5x)\cdot 5=\\\\=6\cdot sin^22x\cdot cos2x-5sin10x\\\\4)\; \; y=ln(x^2+5x+1)\\\\y'=\frac{1}{x^2+5x+1}\cdot(2x+5)\\\\5)\; \; y=(sinx)^{cos2x}\\\\lny=cos2x\cdot ln(sinx)\\\\\frac{y'}{y}=-2sin2x\cdot ln(sinx)+cos2x\cdot \frac{cosx}{sinx}=-2sin2x\cdot ln(sinx)+cos2x\cdot tgx\\\\6)\; \; \left \{ {{y=t^2+2t+1} \atop {x=t+1}} \right. \; \; ,\; \quad t=x-1\\\\y'_{x}=\frac{y'_{t}}{x'_{t}}=\frac{2t+2}{1}=2t+2=2(t+1)=2(x-1+1)=2x$