Partial Derivatives of x^2 sin(y)

One factor moves, the other is a number

Hold $y$ fixed. Then $\sin y$ is some number $s$, and $f = s x^2$ is a plain parabola in $x$ with derivative $2sx$. Restore $s = \sin y$.

Hold $x$ fixed instead. Now $x^2$ is a number and $f$ is a sine wave in $y$ scaled by that number, so the derivative is $x^2\cos y$.

Read them at $\left(1, \frac{\pi}{2}\right)$: $\frac{\partial f}{\partial x} = 2\sin\frac{\pi}{2} = 2$ and $\frac{\partial f}{\partial y} = \cos\frac{\pi}{2} = 0$. The surface is at a crest in the $y$ direction there, so moving in $y$ costs nothing to first order.

The mistake: mixing the two rules together

A frequent wrong answer is $2x\cos y$, which differentiates both factors in the same step. Another is $2x\sin y + x^2\cos y$, which is the full product rule as if $x$ and $y$ were the same variable. Both come from not deciding which variable is moving.

• $2x\sin y$ differentiates $x^2$ only, correct for $\frac{\partial f}{\partial x}$.
• $x^2\cos y$ differentiates $\sin y$ only, correct for $\frac{\partial f}{\partial y}$.
• The sum of the two is the rate of change along the direction $(1,1)$, not a partial derivative at all.

If you ever need that sum, it has a name: it is $\nabla f\cdot(1,1)$, the rate at which $f$ changes along a path that moves one unit in $y$ for every unit in $x$. Divide it by $\sqrt{2}$ if you want the directional derivative in the unit direction, since $(1,1)$ has length $\sqrt{2}$.

Second derivatives and the shape

Differentiate again to see the curvature. $\frac{\partial^2 f}{\partial x^2} = 2\sin y$ and $\frac{\partial^2 f}{\partial y^2} = -x^2\sin y$. Off the line $x = 0$, wherever $\sin y > 0$ the surface curves upward in $x$ and downward in $y$, so those points sit on saddle shaped patches. On $x = 0$ itself the $y$ curvature is zero, and in fact both partials vanish along that whole line, so the second derivative test says nothing there.

The mixed partial is $\frac{\partial^2 f}{\partial x \partial y} = 2x\cos y$ whichever order you take, since differentiating $2x\sin y$ by $y$ and $x^2\cos y$ by $x$ both land there.