$f: \mathbb{R}^2 \to \mathbb{R}$ with $f(x,y) := x^2 \exp y + \exp(xy)$ derivation

Question

How can one give the set of points $D$, in which this function is partially derivable and calculate its partial derivations and gradient there?

$f: \mathbb{R}^2 \to \mathbb{R}$ with $f(x,y) := x^2 \exp y + \exp(xy)$

Can I write $D(f) =$ {$(x,y) | x,y \in \mathbb{R}$}?

So

$$\frac{\partial f}{\partial x} = 2e^{x^2} + ye^{xy}$$

and

$$\frac{\partial f}{\partial y} = xe^{xy}$$

Gradient: $$f = { 2e^{x^2} + ye^{xy} \choose xe^{xy}}$$

Is that wrong or correct?

Answer 1

Both are wrong. You have to treat $e^y$ as a multiplicative constant when you take the derivation in respect to $x$, so $\frac{\partial} {\partial x}(x^2e^y)=2xe^y$, hence the partial derivative of $f$ in respect to $x$ is $2xe^y+ye^{xy}$.

To derivate $f$ in respect to $y$ you have to treat $x$ as a (again) multiplicative constant, so the derivative of $f$ in respect to $y$ is $x^2e^y+xe^{xy}$.