How do you differentiate sqrt(e^(x-y^2)-(xy)^2exy2(xy)2?

1 Answer
Jul 28, 2016

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)))f(x,y)=⎜ ⎜exy22xy22exy2(xy)2,2yexy22x2y2exy2(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))f(x,y)=(fx,fx)

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))fx=exy22xy22exy2(xy)2

(delf)/(dely) = (-2ye^(x-y^2) - 2x^2y)/ (2 sqrt(e^(x-y^2) - (xy)^2))fy=2yexy22x2y2exy2(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)))f(x,y)=⎜ ⎜exy22xy22exy2(xy)2,2yexy22x2y2exy2(xy)2⎟ ⎟