Answer
$\dfrac{\partial f}{\partial u}=2 e^{2u+2v}+2u+2v$
Work Step by Step
We have $\dfrac{\partial f}{\partial x}=2x; \dfrac{\partial f}{\partial y}=2y; \dfrac{\partial x}{\partial u}=e^{u+v}; \dfrac{\partial y}{\partial u}=1$
We apply the chain rule to obtain:
$\dfrac{\partial f}{\partial u}=\dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial u}+\dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial u} \\=2x (e^{u+v})+2y(1)$
We are given that $x=e^{u+v}; y=u+v$
Thus, $\dfrac{\partial f}{\partial u}=2 e^{u+v} \ e^{u+v}+2 (u+v)$
Therefore, we get $\dfrac{\partial f}{\partial u}=2 e^{2u+2v}+2u+2v$
