Answer
$\dfrac{\partial P}{\partial x} =\dfrac{6}{\sqrt 5} \\ \dfrac{\partial P}{\partial y} =\dfrac{2}{\sqrt 5}$
Work Step by Step
$\dfrac{\partial P}{\partial x} =\dfrac{ue^y+vye^x+wye^{xy}}{\sqrt{u^2+v^2+w^2}}$
Plug the values $x=0; y=2$.
$\dfrac{\partial P}{\partial x} =\dfrac{ue^y+vye^x+wye^{xy}}{\sqrt{u^2+v^2+w^2}}=\dfrac{(0)(e^2)}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(2)(2)(e^0)}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(1) \times 2 \times e^0}{\sqrt{(0)^2+(1)^2+(2)^2}}=\dfrac{(0)(e^2)}{\sqrt{5}}+\dfrac{(2)(2)(e^0)}{\sqrt{5}}+\dfrac{(1) \times 2 \times e^0}{\sqrt 5}=\dfrac{6}{\sqrt 5}$;
and
$\dfrac{\partial P}{\partial y} =\dfrac{uxe^y+ve^x+wxe^{xy}}{\sqrt{u^2+v^2+w^2}}$
Plug the values $x=0; y=2$.
$\dfrac{\partial P}{\partial y} =\dfrac{uxe^y+ve^x+wxe^{xy}}{\sqrt{u^2+v^2+w^2}}=\dfrac{(0) \times 0 \times e^2}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(2) (e^0)}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(1) \times (0) \times e^0}{\sqrt{(0)^2+(1)^2+(2)^2}}=\dfrac{(0) \times 0 \times e^2}{\sqrt{5}}+\dfrac{(2) (e^0)}{\sqrt{5}}+\dfrac{(1) \times (0) \times e^0}{\sqrt{5}}=\dfrac{2}{\sqrt 5}$
