Proving Twice Differentiability at a Point for a Function of Two Variables

[nuuskur]

Homework Statement

Given $f(x,y) = x\cdot 3^{x+y^2}$. Prove that f is differentiable twice at the point P(1,0).

Homework Equations

$D\subset\mathbb{R}^2, f\colon D\to\mathbb{R}, P\in \mathring{D}$(interior point) - then f is differentiable n+1 times at P$\Leftrightarrow \exists\varepsilon > 0\colon$ in the sphere $B(P,\varepsilon)$ there are defined n-th degree partial derivatives.

The Attempt at a Solution

I want to use the definition of a partial derivative $f_x := \lim_{t\to 0}\frac{f(P_x + t, P_y) - f(P_x,P_y)}{t}$ if the limit exists (either in R or +/- infinity).
How can I check if the limit exists for any and all points inside a sphere? I want to give an epsilon and show that every point in the sphere with radius epsilon has a final limit for both f_x and f_y.

 

[HallsofIvy]

Have you tried actually writing out the difference quotient for that particular f? Using the fact that $xe^{x+ y^2}= xe^xe^{y^2}$ that should be easy.

 

[nuuskur]

$f_x = 3^{y^2} \lim_{t\to 0} \frac{(x+t)3^x3^t - x3^x}{t}$. This is all kid's play, but how does this help me in determining whether f_x is defined for every point in the epsilon-sphere?

If the limit exists for any x (that is to say, that any f_x is defined in its domain), would it be enough to show that all of the (x,y) in my epsilon sphere (a circle actually) belong to the domain of the function?

Since f is an elementary function that is defined in all of $\mathbb{R}^2$. Therefore any point in the epsilon circle (I keep calling it sphere :/ ) would fit the bill no problem. Would this explanation be sufficient?