Question

Y=(x^(2)-2)/(x+y) Функциянын туындысын табу

Answer 1

$y=\dfrac{x^2-2}{x+y} \ \ ,\ \ \ \ \ \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}\\\\\\y'=\dfrac{2x\, (x+y)-(x^2-2)\cdot (1+y')}{(x+y)^2}\\\\\\(x+y)^2\cdot y'=2x(x+y)-(x^2-2)\cdot (1+y')\\\\y'\cdot \Big((x+y)^2+(x^2-2)\Big)=2x(x+y)-(x^2-2)\\\\\\y'=\dfrac{2x^2+2xy-x^2+2}{2x^2+2xy+y^2-2}$