# How do you differentiate xe^(x^2+y^2)?

Mar 31, 2015

To differentiate a two-variable function, you need to build the gradient. The gradient is a vector with as many coordinates as the variables the function depends on.

Each coordinate of the vector is a derivative with respect to one of the variables. So, in the two-variables case, you need to calculate the derivatives with respect to $x$ and $y$, and then put them together in a vector.

Since deriving with respect to a variable means to consider the other as a constant, it's easier to derivate your function if it's expressed in the form
$x {e}^{{x}^{2} + {y}^{2}} = x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}}$

So, deriving with respect to $x$, and using the product rule $\left(f g\right) ' = f ' g + f g '$ where $f \left(x\right) = x$ and $g \left(x\right) = {e}^{{x}^{2}}$, we get
$\frac{d}{\mathrm{dx}} x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}} = {e}^{{y}^{2}} \left({e}^{{x}^{2}} + x \cdot {e}^{{x}^{2}} \cdot 2 x\right) =$
${e}^{{x}^{2}} \cdot {e}^{{y}^{2}} \left(1 + 2 {x}^{2}\right) = {e}^{{x}^{2} + {y}^{2}} \left(1 + 2 {x}^{2}\right)$
Where the derivative of ${e}^{{x}^{2}}$ has been calculated using the chain rule, which states that $f \left(g \left(x\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$, where $f \left(x\right) = {e}^{x}$, and $g \left(x\right) = {x}^{2}$.

The derivative with respect to $y$ is easier, since the only factor to differentiate is ${e}^{{y}^{2}}$, while the others depend only on $x$ and are thus to be considered as constant. So, we have
$\frac{d}{\mathrm{dy}} x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}} = x \cdot {e}^{{x}^{2}} \cdot {e}^{{y}^{2}} \cdot 2 y =$
$2 x y {e}^{{x}^{2} + {y}^{2}}$

The gradient is thus the vector

$\left({e}^{{x}^{2} + {y}^{2}} \left(1 + 2 {x}^{2}\right) , 2 x y \setminus {e}^{{x}^{2} + {y}^{2}}\right)$