Partial Derivatives of sin(xy)

Outside first, then the inside

Name the inside $u = xy$, so $f = \sin u$. The chain rule for partials says differentiate the outside at the inside, then multiply by the partial derivative of the inside.

With $y$ held fixed, $u = xy$ is a straight line in $x$ with slope $y$, so $\frac{\partial u}{\partial x} = y$. Symmetrically $\frac{\partial u}{\partial y} = x$, which gives the second partial.

At $(2,3)$ the inside is $6$, so $\frac{\partial f}{\partial x} = 3\cos 6 \approx 2.881$ and $\frac{\partial f}{\partial y} = 2\cos 6 \approx 1.920$. Same cosine, different multipliers.

The mistake: forgetting the inside factor

Students write $\frac{\partial f}{\partial x} = \cos(xy)$ and stop. That answer would be correct if the inside were just $x$. Here the inside changes $y$ times as fast as $x$ does, so the rate of change of $f$ is scaled by $y$. Drop that factor and your answer is wrong by a whole variable.

• Write the inside function down explicitly before you differentiate.
• Check units of variation: if $y = 0$ then $f = \sin 0 = 0$ for every $x$, so $\frac{\partial f}{\partial x}$ must be zero along that line, and $y\cos(xy)$ is.
• The bare answer $\cos(xy)$ fails that test, since it equals $1$ when $y = 0$.

A symmetry you can use as a check

Multiply the first partial by $x$ and the second by $y$ and you get the same expression, $xy\cos(xy)$. So the two partials satisfy a single relation at every point.

That is exactly what you expect for a function that depends on $x$ and $y$ only through the product $xy$. Any such function is constant along the hyperbolas $xy = c$, and the relation above is the algebraic fingerprint of that.