Partials of x sin(y) + y cos(x)

Term by term, with the roles swapping

Take $\frac{\partial f}{\partial x}$ first. In the term $x\sin y$ the moving variable sits in the front factor, so its derivative is $\sin y$. In the term $y\cos x$ the moving variable sits inside the cosine, so $y$ is the constant multiplier and you get $-y\sin x$.

Now $\frac{\partial f}{\partial y}$. The roles reverse: $x\sin y$ has $x$ as the multiplier and gives $x\cos y$, while $y\cos x$ has $\cos x$ as the multiplier and gives $\cos x$.

Notice that each partial keeps one piece of each term. That is the signature of a function where the variables genuinely interact.

The mistake: assuming a term drops out

When a term visibly contains the other variable, students often discard it. Here $y\cos x$ contains $y$, but it also contains $x$, so it survives $\frac{\partial f}{\partial x}$ as $-y\sin x$. A term only disappears when the moving variable is completely absent from it.

• Scan each term for the moving variable before deciding anything.
• If the moving variable is present, differentiate that term and keep the rest as a constant multiplier.
• If it is absent, and only then, the term contributes zero.

A second slip is losing the minus sign on $-y\sin x$. It comes from $\frac{d}{dx}\cos x = -\sin x$, not from anything about $y$.

Linearisation at the origin

Evaluate everything at $(0,0)$. The function value is $0\cdot\sin 0 + 0\cdot\cos 0 = 0$. The partials are $\sin 0 - 0\cdot\sin 0 = 0$ and $0\cdot\cos 0 + \cos 0 = 1$.

So near the origin the surface behaves like the plane $z = y$. Test it at $(0.1, 0.1)$: the true value is $0.1\sin(0.1) + 0.1\cos(0.1) \approx 0.1095$, and the linear estimate is $0.1$. The error is about $0.0095$, roughly the size you expect from a quadratic correction at that scale.