Question

what is the value of​

Answer 1

Answer:

1

Step-by-step explanation:

$\displaystyle u=\log\left(\frac{x^2+y^2}{x+y}\right)\\\\\Rightarrow(x+y)e^u=x^2+y^2\\\\\Rightarrow(x+y)e^u\frac{\partial u}{\partial x}+e^u=2x\quad\text{and}\quad(x+y)e^u\frac{\partial u}{\partial y}+e^u=2y$

Adding x times the first of these to y times the second gives:

$\displaystyle(x+y)e^u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)+(x+y)e^u=2(x^2+y^2)\\\\\Rightarrow(x^2+y^2)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)+(x^2+y^2)=2(x^2+y^2)\\\\\Rightarrow\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)+1=2\\\\\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=1$