Multivariable calculus

Partial Derivatives of arctan(xy)

Q: Is the function differentiable everywhere?

Yes. The denominator 1 + x^2y^2 is at least 1 at every point, so both partials are continuous on the whole plane and the function is differentiable there.

Q: Why are the two partials mirror images of each other?

Because f(x,y) = f(y,x). Swapping the variables leaves the function alone, so it must swap the two partials, and indeed \frac{y}{1+x^2y^2} becomes \frac{x}{1+x^2y^2}.

For f(x, y) = arctan(xy), the partial derivative with respect to x is y divided by 1 plus x squared y squared, and the partial derivative with respect to y is x divided by the same denominator. The inverse tangent contributes 1 over 1 plus the inside squared, and the chain rule contributes y or x.

f(x,y) = \arctan(xy)

\frac{\partial f}{\partial x} = \frac{y}{1 + x^2y^2}

\frac{\partial f}{\partial y} = \frac{x}{1 + x^2y^2}

The inverse tangent rule with an inside function

The single variable fact you need is $\frac{d}{du}\arctan u = \frac{1}{1+u^2}$ . With $u = xy$ the denominator becomes $1 + (xy)^2 = 1 + x^2y^2$ , and it is the same for both partials because it depends only on the inside.

\frac{\partial f}{\partial x} = \frac{1}{1+(xy)^2}\cdot\frac{\partial}{\partial x}(xy) = \frac{y}{1+x^2y^2}

\frac{\partial f}{\partial y} = \frac{1}{1+(xy)^2}\cdot\frac{\partial}{\partial y}(xy) = \frac{x}{1+x^2y^2}

At $(1,1)$ both partials equal $\frac{1}{2}$ . The denominator never vanishes, so unlike most quotient answers this one is defined on the whole plane.

The mistake: squaring the wrong thing

The usual slip is writing the denominator as $1 + xy^2$ or $1 + x^2y$ , dropping half of the square. The rule squares the entire inside, so $(xy)^2 = x^2y^2$ and both variables appear squared.

Another version of the error is $1 + x^2 + y^2$ , which comes from squaring the terms separately as if the inside were a sum.
A quick check: on the line $y = 1$ the function is $\arctan x$ , so $\frac{\partial f}{\partial x}$ must reduce to $\frac{1}{1+x^2}$ there. Substituting $y=1$ into $\frac{y}{1+x^2y^2}$ gives exactly that.

What the gradient tells you about the shape

Both partials share a denominator, so $\nabla f = \frac{1}{1+x^2y^2}(y, x)$ . The denominator is positive everywhere, so the direction of steepest ascent is always $(y,x)$ itself. The gradient of $\sin(xy)$ lines up with the same vector, though that one reverses each time $\cos(xy)$ changes sign. Both functions depend on $x$ and $y$ only through the product, which is what fixes the direction.

The size of the gradient tells a different story. Freeze $x = 2$ and walk out in $y$ : the partial $\frac{2}{1+4y^2}$ falls from $0.4$ at $y = 1$ to about $0.0198$ at $y = 5$ . That is the arctangent saturating along that line. Its values are trapped between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ , and a bounded function running out of room has to level off.

That makes $\arctan(xy)$ a standard soft threshold in modelling: hold one input fixed and the other traces an S shaped response that flattens once you push it far enough.

Do not turn that into a claim about the whole surface. On the hyperbola $xy = 10$ the point $\left(\sqrt{10},\sqrt{10}\right)$ has both partials near $0.031$ , but $(100, 0.1)$ sits on the same curve with $\frac{\partial f}{\partial y} = \frac{100}{101} \approx 0.99$ . The product alone does not fix the steepness, because the numerators $y$ and $x$ can be as lopsided as you like.

Frequently asked questions

Is the function differentiable everywhere?

Yes. The denominator $1 + x^2y^2$ is at least $1$ at every point, so both partials are continuous on the whole plane and the function is differentiable there.

Why are the two partials mirror images of each other?

Because $f(x,y) = f(y,x)$ . Swapping the variables leaves the function alone, so it must swap the two partials, and indeed $\frac{y}{1+x^2y^2}$ becomes $\frac{x}{1+x^2y^2}$ .