Partial Derivatives of x arctan(xy)

Product rule outside, arctangent derivative inside

The rule you need is $\frac{d}{du}\arctan u = \frac{1}{1+u^{2}}$, with $u = xy$ here so $u^{2} = x^{2}y^{2}$. In $x$, the function is a product of $x$ and $\arctan(xy)$, and both factors depend on $x$.

In $y$ the leading $x$ is a constant multiplier, so only the arctangent is differentiated. The inner derivative is $x$, and multiplying by the outer constant $x$ produces the square.

Both partials are defined for every $x$ and $y$, since $1 + x^{2}y^{2}$ is at least 1 and the arctangent accepts every real input. That is what makes arctangent a friendlier inverse trig function than arcsine for problems like this.

The mistake: squaring only one variable in the denominator

The single most common wrong denominator here is $1 + xy$ or $1 + x^{2}y$. The arctangent rule squares its entire argument, and the argument is the product $xy$, so the denominator is $1 + (xy)^{2} = 1 + x^{2}y^{2}$.

• Square the whole inside, not part of it: $(xy)^{2} = x^{2}y^{2}$.
• The chain rule factor is the derivative of the inside: $y$ for the $x$-partial and $x$ for the $y$-partial.
• The $x$-partial keeps a bare $\arctan(xy)$ term from the product rule. Losing it is the other frequent error.

Note that $\frac{\partial f}{\partial y}$ is never negative, since both $x^{2}$ and the denominator are non-negative. So the surface never falls as $y$ grows, whatever the sign of $x$. When $x$ is negative that takes two sign flips: the inside $xy$ decreases, so $\arctan(xy)$ decreases, and multiplying by the negative outer $x$ turns the fall back into a rise.

Evaluating where the arctangent is nice

At $(1,1)$ the argument is 1, so $\arctan(1) = \frac{\pi}{4}$ and $1 + x^{2}y^{2} = 2$.

Numerically the $x$-partial is about $1.285$ and the $y$-partial is exactly $0.5$. The gradient leans strongly toward $x$, which is what you expect when $x$ appears both outside and inside the arctangent.