Partial Derivatives of x sin(xy)

Product rule outside, chain rule inside

Differentiating in $x$ means two rules stacked. The outer structure is a product, $x$ times $\sin(xy)$, and the second factor is itself a composition whose inside is $xy$. Do the product rule first, then handle the sine.

With $y$ frozen, the inside $xy$ has $x$-derivative $y$, so the chain rule gives $y\cos(xy)$. Substituting that back in and simplifying gives the answer.

In $y$ the picture is simpler. The leading $x$ is now a constant multiplier, so there is no product to differentiate; only the chain rule acts, and the inside $xy$ has $y$-derivative $x$. That second $x$ is where the square comes from.

The mistake: losing the inner derivative

The most common wrong answer for the $y$-partial is $x\cos(xy)$. That drops the inner derivative and leaves you one factor of $x$ short. Every time the inside of a trig function is a product of variables, the chain rule contributes the other variable.

• Differentiating $\sin(xy)$ in $x$ contributes $y$, not 1.
• Differentiating $\sin(xy)$ in $y$ contributes $x$, not 1.
• The outer $x$ in $f$ is differentiated only in the $x$-partial, never in the $y$-partial.

A second slip is writing the $x$-partial as $\sin(xy) + x\cos(xy)$, which forgets the inner $y$ on the chain rule term. Check the shape of that term: it must carry both an $x$ from the product rule and a $y$ from the inside.

Reading the answer at a point

Take $x = 1$ and $y = \frac{\pi}{2}$. Then $xy = \frac{\pi}{2}$, so $\sin(xy) = 1$ and $\cos(xy) = 0$. The cosine term switches off entirely.

The zero in the $y$ direction is honest information: at that point the sine has hit its peak, so nudging $y$ changes nothing to first order. Moving in $x$ still changes $f$, because the leading factor of $x$ is growing.