Question

Its partial differentiation ​

Answer 1

It looks like you're supposed to show

`u_x+u_y=u`

We have

`u=(e^(x+y))/(e^x+e^y)`

In the numerator, we have
`e^(x+y)=e^xe^y`, so that the derivative wrt either
`x` or
`y` is simply
`e^xe^y=e^(x+y)`. In the denominator, either
`e^x` or
`e^y` vanishes.

Differentiating wrt
`x` gives, by the quotient rule,

`u_x=(e^(x+y)(e^x+e^y)-e^(x+y)e^x)/((e^x+e^y)^2)=(e^(x+y)e^y)/((e^x+e^y)^2)`

Similarly, differentiating wrt
`y` gives

`u_y=(e^(x+y)e^x)/((e^x+e^y)^2)`

Then

`u_x+u_y=\frac{e^(x+y)(e^y+e^x)}(e^x+e^y)^2}`

`u_x+u_y=\frac{e^(x+y)}e^x+e^y}=u`

as required.