Answer
$\dfrac{\partial P}{\partial x} =\dfrac{6}{\sqrt 5}$;
and $\dfrac{\partial P}{\partial y} =\dfrac{2}{\sqrt 5}$;
Work Step by Step
Here, we have $\dfrac{\partial P}{\partial x} =\dfrac{ue^y}{\sqrt{u^2+v^2+w^2}}+\dfrac{vye^x}{\sqrt{u^2+v^2+w^2}}+\dfrac{wye^{xy}}{\sqrt{u^2+v^2+w^2}}$
When $x=0; y=2$, we have
$\dfrac{\partial P}{\partial x} =\dfrac{0 \times e^2}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(2)(2)(e^0)}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(1) \times 2 \times e^0}{\sqrt{(0)^2+(1)^2+(2)^2}}=\dfrac{6}{\sqrt 5}$;
and
$\dfrac{\partial P}{\partial y} =\dfrac{uxe^y}{\sqrt{u^2+v^2+w^2}}+\dfrac{ve^x}{\sqrt{u^2+v^2+w^2}}+\dfrac{wxe^{xy}}{\sqrt{u^2+v^2+w^2}}$
When $x=0; y=2$, we have
$\dfrac{\partial P}{\partial y} =\dfrac{(0) \times 0 \times e^2}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(2) (e^0)}{\sqrt{(0)^2+(1)^2+(2)^2}}+\dfrac{(1) \times (0) \times e^0}{\sqrt{(0)^2+(1)^2+(2)^2}}=\dfrac{2}{\sqrt 5}$;
Hence, we have $\dfrac{\partial P}{\partial x} =\dfrac{6}{\sqrt 5}$;
and $\dfrac{\partial P}{\partial y} =\dfrac{2}{\sqrt 5}$;
