Question

How do you find `(delf(x,y))/(dely)` and `(delf(x,y))/(delx)` of `f(x,y)=(yx^2-2y)/(2ye^x+y^-3)`, using the quotient rule?

Answer 1

`(delf(x,y))/(dely)=((2ye^x+y^-3)(x^2-2)-(yx^2-2y)(2e^x-3y^-4))/(2ye^x+y^-3)^2`

`(delf(x,y))/(delx)=(2yx(2ye^x+y^-3)-2ye^x(yx^2-2y))/(2ye^x+y^-3)^2`

Explanation:

To partially differentiate, we treat one variable as a constant.

So, for `(delf(x,y))/(dely)` we treat `x` as a constant.

`del/(delx)u/v=(v(delu)/(delx)-u(delv)/(delx))/v^2`
`del/(dely)u/v=(v(delu)/(dely)-u(delv)/(dely))/v^2`

`u=yx^2-2y`
`(delu)/(dely)=x^2-2`

`v=2ye^x+y^-3`
`(delv)/(dely)=2e^x-3y^-4`

`(delf(x,y))/(dely)=((2ye^x+y^-3)(x^2-2)-(yx^2-2y)(2e^x-3y^-4))/(2ye^x+y^-3)^2`

Now, for `(delf(x,y))/(delx)` we treat `y` as a constant.

`u=yx^2-2y`
`(delu)/(delx)=2yx`

`v=2ye^x+y^-3`
`(delv)/(delx)=2ye^x`

`(delf(x,y))/(dely)=(2yx(2ye^x+y^-3)-2ye^x(yx^2-2y))/(2ye^x+y^-3)^2`