0 0 votes If $u=\log (e^x+e^y),$ then $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=$ $e^x+e^y$ $e^x-e^y$ $\frac{1}{e^x+e^y}$ $1$ Calculus gate2013 calculus partial-derivatives + – Milicevic3306 7.9k points answer comment Share Follow 0 reply Please log in or register to add a comment.