Partial Derivatives of cos(x^2 + y^2)

Chain rule with a two variable inside

Set $u = x^2 + y^2$. The outside derivative is $\frac{d}{du}\cos u = -\sin u$, and it is the same for both partials. Only the inside derivative changes.

Multiply the two pieces together for each variable.

At $(1,1)$ the inside is $2$, so both partials equal $-2\sin 2 \approx -1.819$. Equal partials are no accident here: the point sits on the diagonal, and the surface is rotationally symmetric.

The mistake: losing the inside derivative or the sign

Two errors dominate. The first is writing $-\sin(x^2 + y^2)$ with no factor of $2x$, which forgets that the inside is not simply $x$. The second is writing $2x\sin(x^2+y^2)$ with the sign of the cosine derivative dropped.

• Test the sign at $x = 0.5$, $y = 0$: the inside is $0.25$, $\sin(0.25) > 0$, so $\frac{\partial f}{\partial x} = -1\cdot\sin(0.25) < 0$. The ripple is heading downhill away from the central crest, which matches.
• Test the missing factor along the $y$ axis: at $x = 0$ the function is even in $x$, so $\frac{\partial f}{\partial x}$ must be zero, and only the version carrying $2x$ gives that.

The gradient lies along the radius

Collect the two partials into one vector.

The gradient is always a scalar multiple of the position vector $(x,y)$, so it is radial. The sign of that multiple flips with the sine: the gradient points straight out from the origin where $\sin(x^2+y^2) < 0$ and straight back in where the sine is positive. That is the analytic version of the statement that the level curves are circles $x^2 + y^2 = c$, and the gradient is perpendicular to them.

One consequence is worth memorising as a check: $y\frac{\partial f}{\partial x} - x\frac{\partial f}{\partial y} = 0$ everywhere, because both partials share the same sine factor and differ only by $x$ against $y$. Any function of $x^2 + y^2$ alone satisfies that identity.