Question

Помогите решить, очень нужно.

Answer 1

$z=sqrt{2x+y^2-1};, frac{partial z}{partial x}= frac{1}{sqrt{2x+y^2-1}};, frac{partial z}{partial y}= frac{y}{sqrt{2x+y^2-1}};,$
$frac{partial^2 z}{partial x^2}= frac{partial}{partial x} (frac{1}{sqrt{2x+y^2-1}})=-frac{1}{sqrt{(2x+y^2-1)^3}};, frac{partial^2 z}{partial x^2}(1;0)=-1;$
$frac{partial^2 z}{partial y^2}=frac{sqrt{2x+y^2-1}-frac{y^2}{sqrt{2x+y^2-1}}}{2x+y^2-1}=frac{2x-1}{sqrt{(2x+y^2-1)^3}};$
$frac{partial^2 z}{partial y^2}(1;0)=1;,frac{partial^2 z}{partial y partial x}= frac{partial}{partial y} (frac{1}{sqrt{2x+y^2-1}})=-frac{y}{sqrt{(2x+y^2-1)^3}};$
$frac{partial^2 z}{partial x partial y}= frac{partial}{partial x} (frac{y}{sqrt{2x+y^2-1}})=-frac{y}{sqrt{(2x+y^2-1)^3}};$
$P(x, y)=2cdot(-1)-3=-5.$