Partial derivation of $xy \log (1+ \sqrt{x^2 + 2y^2})$

Question

Let $p: \mathbb{R^2} \to \mathbb{R}$ with $p(x,y):= xy \log (1+ \sqrt{x^2 + 2y^2}) $

I want to find the set of points $D$, in which this function is partially differentiable and calculate its partial derivatives and gradient there.

So I would write:

$D(p)=\{(x,y) \mid x,y \in \mathbb{R^{+}} $}

$$\frac{\partial p}{\partial x} = xy \log (1+ \sqrt{x^2 + 2y^2}) $$ When I tried to get the partial function in the internet, I got the following:

and

$$\frac{\partial p}{\partial y} = xy \log (1+ \sqrt{x^2 + 2y^2})$$ here I got that:

I don't understand how though. Can someone explain how to get the partial derivative here?

Answer 1

The partial derivative with respect to $x$ is computed by keeping $y$ constant; so you get $$ \frac{\partial f}{\partial x}=y\log(1+\sqrt{x^2+2y^2})+xy\frac{\dfrac{x}{\sqrt{x^2+2y^2}}}{1+\sqrt{x^2+2y^2}} $$ with issues only at $(0,0)$, which you can solve yourself.

Similarly for the other partial derivative.