Partial Derivatives of xy^2/(1 + x^2)

Split the function before differentiating

Write $f = \frac{x}{1+x^2} \cdot y^2$. The two factors depend on different variables, which decides which rule each partial needs.

For $\frac{\partial f}{\partial y}$, the first factor is a constant multiplier, so only $y^2$ is differentiated.

For $\frac{\partial f}{\partial x}$, pull $y^2$ out front as the constant and use the quotient rule on $\frac{x}{1+x^2}$.

At $(1,2)$ the $x$ partial is $\frac{4(1-1)}{4} = 0$ and the $y$ partial is $\frac{2(1)(2)}{2} = 2$.

The mistake: quotient rule on the wrong variable

Applying the quotient rule to $\frac{\partial f}{\partial y}$ produces the numerator $2xy\left(1+x^2\right) - xy^2(2x)$, treating the denominator as if it changed with $y$. It does not, and the extra term is pure error.

• Before differentiating, ask which factors actually contain your variable.
• $1 + x^2$ has no $y$, so its $y$ derivative is $0$ and the quotient rule degenerates to a constant multiple.
• $y^2$ has no $x$, so it factors straight out of the $x$ derivative.

The other slip is inside the quotient rule itself: the numerator is $u'v - uv'$ in that order. Reversing it flips the sign and turns $1 - x^2$ into $x^2 - 1$.

Where the surface is flat in x

The factor $1 - x^2$ makes $\frac{\partial f}{\partial x}$ vanish along the vertical lines $x = 1$ and $x = -1$, and also along $y = 0$.

That matches the one-variable picture: for fixed $y \neq 0$, the slice $g(x) = \frac{xy^2}{1+x^2}$ has a maximum at $x = 1$ and a minimum at $x = -1$, the classic shape of $\frac{x}{1+x^2}$.

Both partials vanish together only when $y = 0$, so the whole $x$ axis consists of critical points. Along that line the surface is flat, and $f = 0$ there.