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

Chain rule with a saddle in the exponent

The exponent $u = x^{2}-y^{2}$ is the standard saddle. Hold $y$ fixed and $\frac{\partial u}{\partial x} = 2x$; hold $x$ fixed and $\frac{\partial u}{\partial y} = -2y$. In both cases the exponential itself is copied down unchanged and multiplied by that inside derivative.

At $(1,1)$ the exponent is $0$, so $f = 1$, and the slopes are $2$ and $-2$. Stepping in the positive $x$ direction raises the surface; stepping in the positive $y$ direction lowers it by the same amount.

The mistake: a dropped sign or a mishandled exponent

• Writing $2y\,e^{x^{2}-y^{2}}$ for the $y$ partial keeps the size but loses the sign. The exponent falls as $|y|$ grows, so for $y > 0$ the slope in $y$ is negative.
• Writing $(2x-2y)e^{x^{2}-y^{2}}$ differentiates the whole exponent in a single partial. Each partial sees only its own variable move.
• Writing $x^{2}e^{x^{2}-y^{2}}$ copies the exponent down in front instead of its derivative, and writing $e^{2x}$ pushes the inside derivative up into the exponent. The inside derivative is $2x$, and it multiplies the exponential; the exponential itself is copied unchanged.

Use the axes as a check. Along $y = 0$ the function is $e^{x^{2}}$, whose slope is $2xe^{x^{2}}$, and along $x = 0$ it is $e^{-y^{2}}$, whose slope is $-2ye^{-y^{2}}$. Both agree with the general formulas, and a candidate answer that fails these one-variable cases is wrong.

The origin is still a saddle

The exponential is never zero, so the gradient vanishes only where $2x = 0$ and $-2y = 0$, that is, only at the origin, where $f(0,0) = 1$.

That point is a saddle. Along the $x$-axis the function is $e^{x^{2}} \ge 1$, a valley bottom that rises away from the origin; along the $y$-axis it is $e^{-y^{2}} \le 1$, a ridge that falls away. Increasing in one direction and decreasing in the other at the same point is exactly what a saddle means.

This is a useful general fact: composing a function with an increasing outer function such as $e^{u}$ changes the heights but not the pattern of increase and decrease, so the saddle of $x^{2}-y^{2}$ survives the exponential intact.