Question

помогите решить е) и ж) срочно !!!!!!!!!!

Answer 1

$z = {x}^{3} + {y}^{3} - 3xy \ \ frac{dz}{dx} = 3 {x}^{2} + 0 - 3y = 3 {x}^{2} - 3y = 3( {x}^{2} - y) \ \ frac{dz}{dy} = 3 {y}^{2} - 3x = 3( {y}^{2} - x)$
$z = {e}^{ - {x}^{3} } tg(4x) \ \ frac{dz}{dx} = {e}^{ { - x}^{3} } times ( - 3 {x}^{2} ) times tg(4x) + {e}^{ { - x}^{3} } times ( { frac{1}{cos(4x)} }) ^{2} times 4 = \ \ = frac{ - 3 {x}^{2} times cos(4x) times sin(4x) + 4 }{ {e}^{ {x}^{3} } times {cos}^{2} (4x) } = frac{ - 3 {x}^{2} sin(8x) + 8 }{2 {e}^{ {x}^{3} } {cos}^{2} (4x)} \ \ frac{dz}{dy} = 0$


Вот.