Partial Derivatives of e^(x/(1+y^2))

Differentiate the exponent, then multiply

Name the exponent $g(x,y) = \frac{x}{1+y^{2}}$. Every partial of $f$ is $e^{g}$ times the matching partial of $g$, so the only real work is differentiating $g$. When $x$ moves, $g$ is a constant multiple of $x$.

When $y$ moves, write $g = x\left(1+y^{2}\right)^{-1}$ and use the power rule with the chain rule: the outside gives $-\left(1+y^{2}\right)^{-2}$ and the inside gives $2y$.

At $(2,1)$ the exponent is $1$, so $f = e \approx 2.718$, the $x$ slope is $\frac{e}{2} \approx 1.359$, and the $y$ slope is $-e \approx -2.718$.

The mistake: mishandling the fraction in the exponent

• Writing $\frac{\partial g}{\partial y} = \frac{x}{2y}$ differentiates the denominator where it stands, as though the derivative of $\frac{x}{u}$ were $x$ over the derivative of $u$. The reciprocal $\left(1+y^{2}\right)^{-1}$ differentiates to $-2y\left(1+y^{2}\right)^{-2}$, so the whole factor changes, not just the bottom.
• Reporting only the exponent's derivative and dropping the factor $e^{g}$. The exponential is never lost in a chain rule; it is copied down.
• Losing the minus sign from the power $-1$, which would make the surface appear to rise in $y$ where it actually falls.

Check on the $y$-axis. When $x = 0$ the exponent is $0$ for every $y$, so $f$ is identically $1$ along that line and the slope in $y$ must be $0$. The correct formula carries a factor of $x$ and vanishes there, which a version missing that factor would not.

The ridge along the x-axis

For a fixed $x > 0$, the exponent $\frac{x}{1+y^{2}}$ is largest at $y = 0$ and shrinks toward $0$ as $|y|$ grows. So each slice is a bump of height $e^{x}$ on the $x$-axis that decays to height $1$ far away, and the surface has a ridge running along the $x$-axis.

The $y$ partial confirms it. For $x > 0$ its sign is the sign of $-y$, so the surface climbs toward the axis from below and falls away above it, and the slope is exactly $0$ on the ridge line $y = 0$.

The denominator $1+y^{2}$ is never zero, so both formulas hold on the whole plane, and for $x > 0$ the value stays between $1$ and $e^{x}$.