Question

дифференциал функции​

Answer 1

$\boxed {\; dy=y'(x)\cdot dx\; }\\\\\\1)\; \; y=3x^2-4x\\\\dy=(6x-4)\, dx\\\\\\2)\; \; y=x+\sqrt{x^3+2x}\\\\y'=1+\frac{1}{2\sqrt{x^3+2x}}\cdot (3x^2+2)\\\\dy=\Big (1+\frac{3x^2+2}{2\sqrt{x^3+2x}}\Big )\cdot dx\\\\\\3)\; \; y=\frac{x^2+x-3}{x+5}\\\\y'=\frac{(2x+1)(x+5)-(x^2+x-3)}{(x+5)^2}=\frac{2x^2+11x+5-x^2-x+3}{(x+5)^2}=\frac{x^2+10x+8}{(x+5)^2}\\\\dy=\frac{x^2+10x+8}{(x+5)^2}\cdot dx$