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