# First discribe the domain of the given function then find partial derivatives fx, fy,fxx,fxy f(x,y)=e^(2x-y) + ln(y^(2) - 2x)Thanks

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### 1 Answer

Notice that the logarithm exists for `y^2 - 2x gt 0,` hence, x needs to be larger than 0.

Hence, the domain of the function is `[0,+oo).`

You need to find the first order partial derivative f_x, hence, you need to differentiate the given function with respect to x, considering y as constant such that:

`f_x= (del(e^(2x-y) + ln(y^2 - 2x)))/(del x)`

`f_x = 2e^(2x-y)- 2/(y^2 - 2x)`

You need to find the first order partial derivative `f_y` , hence, you need to differentiate the given function with respect to y, considering x as constant such that:

`f_y = (del(e^(2x-y) + ln(y^2 - 2x)))/(del y)`

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

You need to evaluate the second order partial derivative `f_(x x)` such that:

`f_(x x) = (del(2e^(2x-y) - 2/(y^2 - 2x)))/(del x)`

`f_(x x) = 4e^(2x-y)- 4/((y^2 - 2x)^2)`

`f_(xy) = (del(2e^(2x-y) - 2/(y^2 - 2x)))/(del y)`

`f_(x y) = -2e^(2x-y) + (4y)/((y^2 - 2x)^2)`

**Hence, evaluating the first and second order partial derivatives yields `f_x = 2e^(2x-y) - 2/(y^2 - 2x) ; f_y = -e^(2x-y) + (2y)/(y^2 - 2x) ; f_(x y) = -2e^(2x-y) + (4y)/((y^2 - 2x)^2).` **