Partial Derivatives of xy e^(-x^2 - y^2)

Product rule with a Gaussian factor

Each variable appears twice: once in the polynomial factor $xy$ and once in the exponent. So both partials need the product rule, and in each case the exponential contributes its own chain rule factor.

Differentiating in $x$, treat $y$ as a constant. The polynomial part contributes $y$, and the exponential contributes $-2x$ times itself.

The exponential is common to both terms, so factor it out along with the $y$. What is left inside the bracket is the informative part.

The function is symmetric in $x$ and $y$, so the $y$-partial is the same expression with the letters swapped.

The mistake: dropping a term or a sign in the exponent

Two errors are worth naming. The first is writing the $x$-partial as $-2x^{2}y\,e^{-x^{2}-y^{2}}$, which keeps only the second product rule term. The second is losing the minus sign on the chain rule factor and getting $1 + 2x^{2}$ inside the bracket, which never vanishes and so destroys the critical point structure.

• The exponent is $-x^{2}-y^{2}$, so its $x$-derivative is $-2x$ and its $y$-derivative is $-2y$.
• The bracket $1 - 2x^{2}$ is zero at $x = \pm\frac{1}{\sqrt{2}}$, which is the $x$-coordinate of the four peaks and valleys, not a peak on its own.
• The exponential factor is positive everywhere, so it never creates a zero of its own.

Setting both partials to zero gives $y(1-2x^{2}) = 0$ and $x(1-2y^{2}) = 0$. The solutions are the origin, which is a saddle, and the four points where $x$ and $y$ are each $\pm\frac{1}{\sqrt{2}}$, which are the two peaks and two valleys of this classic bump surface.

A worked evaluation

At $(1,1)$ the exponent is $-2$, and the bracket $1 - 2(1)^{2} = -1$ for both partials.

Both components are about $-0.135$. The point $(1,1)$ sits outside the peak at $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, so moving further out in either direction sends you downhill, which is exactly what the negative signs say.