Question

Помогите пожалуйста!! Срочно ​

Answer 1

Ответ:

$1,6$

Объяснение:

$g(x,y)=\dfrac{4x}{x-2y};$

$\overrightarrow {a}= \{-4; -3\};$

$\dfrac{\partial g}{\partial a}=\dfrac{\partial g}{\partial x} \bigg |_{(0; 1)} \cdot cos\alpha+\dfrac{\partial g}{\partial y} \bigg |_{(0; 1)} \cdot cos\beta;$

$\dfrac{\partial g}{\partial x}=\dfrac{(4x)_{x}^{'} \cdot (x-2y)-4x \cdot (x-2y)_{x}^{'}}{(x-2y)^{2}}=\dfrac{4(x-2y)-4x}{(x-2y)^{2}}=\dfrac{4x-8y-4x}{(x-2y)^{2}}=$

$=\dfrac{-8y}{(x-2y)^{2}};$

$\dfrac{\partial g}{\partial y}=\dfrac{(4x)_{y}^{'} \cdot (x-2y)-4x \cdot (x-2y)_{y}^{'}}{(x-2y)^{2}}=\dfrac{0-4x \cdot (-2)}{(x-2y)^{2}}=\dfrac{8x}{(x-2y)^{2}};$

$\dfrac{\partial g}{\partial x} \bigg |_{(0; 1)}^{}=\dfrac{-8 \cdot 1}{(0-2 \cdot 1)^{2}}=\dfrac{-8}{4}=-2;$

$\dfrac{\partial g}{\partial y} \bigg |_{(0;1)}^{}=\dfrac{8 \cdot 0}{(0-2 \cdot 1)^{2}}=0;$

$cos\alpha=\dfrac{a_{x}}{|\overrightarrow {a}|} \quad , \quad cos\beta=\dfrac{a_{y}}{|\overrightarrow {a}|} \quad ;$

$cos\alpha=\dfrac{-4}{\sqrt{(-4)^{2}+(-3)^{2}}}=\dfrac{-4}{\sqrt{16+9}}=\dfrac{-4}{\sqrt{25}}=\dfrac{-4}{5}=-0,8;$

$cos\beta=\dfrac{-3}{\sqrt{(-4)^{2}+(-3)^{2}}}=\dfrac{-3}{5}=-0,6;$

$\dfrac{\partial g}{\partial a}=-2 \cdot (-0,8)+0 \cdot (-0,6)=1,6;$