Critical points, extrema, and saddle points of $$$f{\left(x,y \right)} = e^{x y}$$$

Find and classify the critical points of $$$f{\left(x,y \right)} = e^{x y}$$$.

The first step is to find all the first-order partial derivatives:

$$$\frac{\partial}{\partial x} \left(e^{x y}\right) = y e^{x y}$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial y} \left(e^{x y}\right) = x e^{x y}$$$ (for steps, see partial derivative calculator).

Next, solve the system $$$\begin{cases} \frac{\partial f}{\partial x} = 0 \\ \frac{\partial f}{\partial y} = 0 \end{cases}$$$, or $$$\begin{cases} y e^{x y} = 0 \\ x e^{x y} = 0 \end{cases}$$$.

The system has the following real solution: $$$\left(x, y\right) = \left(0, 0\right)$$$.

Now, let's try to classify it.

Find all the second-order partial derivatives:

$$$\frac{\partial^{2}}{\partial x^{2}} \left(e^{x y}\right) = y^{2} e^{x y}$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial^{2}}{\partial y\partial x} \left(e^{x y}\right) = \left(x y + 1\right) e^{x y}$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial^{2}}{\partial y^{2}} \left(e^{x y}\right) = x^{2} e^{x y}$$$ (for steps, see partial derivative calculator).

Define the expression $$$D = \frac{\partial ^{2}f}{\partial x^{2}} \frac{\partial ^{2}f}{\partial y^{2}} - \left(\frac{\partial ^{2}f}{\partial y\partial x}\right)^{2} = - \left(2 x y + 1\right) e^{2 x y}.$$$

Since $$$D{\left(0,0 \right)} = -1$$$ is less than $$$0$$$, it can be stated that $$$\left(0, 0\right)$$$ is a saddle point.

Relative Maxima

No relative maxima.

Relative Minima

No relative minima.

Saddle Points

$$$\left(x, y\right) = \left(0, 0\right)$$$A, $$$f{\left(0,0 \right)} = 1$$$A