Question

решите,пожалуйста))))

Answer 1

$1)\; \; y= \frac{1-x^2}{1+x^2}\\\\y'=\frac{-2x(1+x^2)-(1-x^2)\cdot 2x}{(1+x^2)^2}=\frac{-2x-2x^3-2x+2x^3}{(1+x^2)^2}=-\frac{4x}{(1+x^2)^2}\\\\2)\; \; y=sin(x^2+1)\\\\y'=cos(x^2+1)(x^2+1)'=cos(x^2+1)\cdot 2x\\\\y''=-sin(x^2+1)\cdot \underbrace {(x^2+1)'}_{2x}\cdot 2x+cos(x^2+1)\cdot 2=\\\\=-4x\cdot sin(x^2+1)+2\cdot cos(x^2+1)\\\\3)\; \; y=x^2-8x+12\\\\y'=2x-8\qquad znaki\; y':\; ---\underbrace {(4)}_{y'(4)=0}+++\\\\2x-8\ \textgreater \ 0\\\\2x\ \textgreater \ 8\\\\x\ \textgreater \ 4\\\\y(x)\; \; vozrastaet\; pri\; x\in [\, 4,+\infty )$