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

Jul 28, 2016

$\nabla f \left(x , y\right) = \left(\frac{{e}^{x - {y}^{2}} - 2 x {y}^{2}}{2 \sqrt{{e}^{x - {y}^{2}} - {\left(x y\right)}^{2}}} , \frac{- 2 y {e}^{x - {y}^{2}} - 2 {x}^{2} y}{2 \sqrt{{e}^{x - {y}^{2}} - {\left(x y\right)}^{2}}}\right)$

#### 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:

$\nabla f \left(x , y\right) = \left(\frac{\partial f}{\partial x} , \frac{\partial f}{\partial x}\right)$

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

$\frac{\partial f}{\partial x} = \frac{{e}^{x - {y}^{2}} - 2 x {y}^{2}}{2 \sqrt{{e}^{x - {y}^{2}} - {\left(x y\right)}^{2}}}$

$\frac{\partial f}{\partial y} = \frac{- 2 y {e}^{x - {y}^{2}} - 2 {x}^{2} y}{2 \sqrt{{e}^{x - {y}^{2}} - {\left(x y\right)}^{2}}}$

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

$\nabla f \left(x , y\right) = \left(\frac{{e}^{x - {y}^{2}} - 2 x {y}^{2}}{2 \sqrt{{e}^{x - {y}^{2}} - {\left(x y\right)}^{2}}} , \frac{- 2 y {e}^{x - {y}^{2}} - 2 {x}^{2} y}{2 \sqrt{{e}^{x - {y}^{2}} - {\left(x y\right)}^{2}}}\right)$