Partial Derivatives of sin(xyz): Chain Rule

A triple product, one variable at a time

Set $u = xyz$. Holding $y$ and $z$ fixed makes $u$ a constant multiple of $x$, so $\frac{\partial u}{\partial x} = yz$. The same reasoning gives $xz$ and $xy$ for the other two variables.

The rule of thumb: the coefficient of each partial is the product of the variables you did not differentiate with respect to. That is the constant-multiple rule doing its job, not the product rule.

The mistake: using the product rule on xyz

A partial derivative freezes the other variables, so $xyz$ is not a product of three functions of $x$. Applying the product rule anyway and writing $yz + xz + xy$ for $\frac{\partial u}{\partial x}$ answers a different question. That sum is the rate of change when all three variables increase together at the same rate, a directional derivative along $(1,1,1)$, not a partial derivative at all.

Test it at $(1, 2, 3)$, where $u = 6$. The true $f_x = 2\cdot 3\cos 6 = 6\cos 6 \approx 5.761$. The product rule answer uses $2\cdot3 + 1\cdot3 + 1\cdot2 = 11$ and gives about $10.562$, nearly double.

• Freeze the other two variables and name their product.
• Differentiate the outer sine, leaving $xyz$ intact inside the cosine.
• Multiply by the frozen product, not by a sum of three terms.

Symmetry as a free check

The function is unchanged if you swap any two of its variables, so its partials must respect that too. Swap $y$ and $z$ in $f_x = yz\cos(xyz)$ and the expression comes back identical. A candidate answer that breaks this symmetry is wrong before you test a single number.

On the plane $x = 0$ the product $xyz$ is zero, so $\cos(xyz) = 1$ and the three partials collapse to $yz$, $0$ and $0$. The gradient there points purely along the $x$ direction, which says the function changes only as you step off that plane.