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

Run the quotient rule with the other letters frozen

Here the differentiating variable appears both upstairs and downstairs, so the quotient rule is genuinely needed. Call the numerator $N = xyz$ and the denominator $D = 1 + x^2 + y^2 + z^2$. With $y$ and $z$ held fixed, $\frac{\partial N}{\partial x} = yz$ and $\frac{\partial D}{\partial x} = 2x$.

Now factor $yz$ out of the numerator. The two $x^2$ terms combine as $x^2 - 2x^2 = -x^2$, which is where the single minus sign comes from.

The function is symmetric in the three variables, so the other two answers come from relabelling rather than from repeating the whole calculation. In each one, the coefficient out front is the product of the two variables you held fixed, and the minus sign sits on the square of the variable you differentiated.

The mistake: the order of the two terms on top

The quotient rule is not symmetric, so writing the numerator backwards flips the sign of the entire answer. The order is denominator times derivative of numerator, minus numerator times derivative of denominator. Many students find it easier to remember as low times high derivative, minus high times low derivative, all over low squared.

The second frequent error is forgetting to square the denominator, leaving $D$ instead of $D^2$. Check the size of the terms: the numerator of the answer has degree four, so an unsquared denominator of degree two would make the partial derivative grow without bound. With $D^2$ the degrees match, four against four, so the answer levels off instead: along the ray $x = 0$ with $y = z$ growing, $\frac{\partial f}{\partial x}$ approaches $\frac{1}{2}$.

• The denominator of the answer is $D^2$, never $D$.
• The minus sign belongs to the term carrying the derivative of the denominator.
• After factoring, exactly one square inside the bracket is negative, and it is the one matching the variable you differentiated.

A sample point and a sanity check on the pattern

At $(1,1,1)$ the denominator is $4$, so $f = \frac{1}{4}$. The $x$ partial is $\frac{1 \cdot (1 - 1 + 1 + 1)}{16} = \frac{2}{16} = \frac{1}{8}$, and by symmetry the other two are also $\frac{1}{8}$.

Two checks are cheap and catch most errors. On the plane $x = 0$ the $x$ partial reduces to $\frac{yz(1+y^2+z^2)}{(1+y^2+z^2)^2}$, which simplifies to $\frac{yz}{1+y^2+z^2}$, exactly what you get by differentiating the original function along that plane. And if either $y$ or $z$ is zero then the $x$ partial is zero, which it must be, because setting $y = 0$ makes $f$ identically zero for all $x$.