Question

How do you differentiate `sqrt(e^(x-y^2)-(xy)^2`?

Answer 1

`grad f(x,y) = ((e^(x-y^2) - 2xy^2)/ (2 sqrt(e^(x-y^2) - (xy)^2)) , (-2ye^(x-y^2) - 2x^2y)/ (2 sqrt(e^(x-y^2) - (xy)^2)))`

Explanation:

You've presented a three dimensional function for differentiation. The common method of presenting a "derivative" for such a function is to use the gradient:

`grad f(x,y) = ((delf)/(delx) , (delf)/(delx))`

So we'll compute each partial individually and the result will be the gradient vector. Each can be easily determined using the chain rule.

`(delf)/(delx) = (e^(x-y^2) - 2xy^2)/ (2 sqrt(e^(x-y^2) - (xy)^2))`

`(delf)/(dely) = (-2ye^(x-y^2) - 2x^2y)/ (2 sqrt(e^(x-y^2) - (xy)^2))`

From here, denoting the gradient is as easy as incorporating these into the gradient vector:

`grad f(x,y) = ((e^(x-y^2) - 2xy^2)/ (2 sqrt(e^(x-y^2) - (xy)^2)) , (-2ye^(x-y^2) - 2x^2y)/ (2 sqrt(e^(x-y^2) - (xy)^2)))`