Partial Derivatives of sin(x + 2y)

A linear inside makes the chain rule easy

Let $u = x + 2y$. Since $u$ is linear, its partials are constants: $\frac{\partial u}{\partial x} = 1$ and $\frac{\partial u}{\partial y} = 2$. The outside derivative is $\cos u$ in both cases.

So $\frac{\partial f}{\partial y}$ is exactly twice $\frac{\partial f}{\partial x}$ at every point. At the origin they are $1$ and $2$. Moving in $y$ changes the phase twice as fast as moving in $x$, so the surface is twice as steep in that direction.

The mistake: putting the 2 in the wrong place

The two wrong answers you see most are $\cos(x + 2y)\cdot 2y$, which differentiates the inside as if it were a product, and $\cos(2x + 2y)$, which moves the coefficient inside the cosine. The inside function is not touched by differentiation, only evaluated.

• The inside stays $x + 2y$ in every answer. Only the multiplier out front can change.
• The multiplier is the partial derivative of the inside, and $\frac{\partial}{\partial y}(x + 2y) = 2$, not $2y$.
• Sanity check by restricting to $x = 0$: the function is $\sin(2y)$, whose derivative is $2\cos(2y)$, and setting $x = 0$ in $2\cos(x+2y)$ gives the same.

A wave moving through the plane

The level curves are the lines $x + 2y = c$, and the function is a wave crest sliding along the direction $(1,2)$. Differentiating twice gives $\frac{\partial^2 f}{\partial x^2} = -\sin(x+2y)$ and $\frac{\partial^2 f}{\partial y^2} = -4\sin(x+2y)$, so the second derivatives are locked together.

Reading $y$ as time, this is the wave equation with speed $2$: a crest sits where $x + 2y = c$, so each unit of $y$ moves it two units in the negative $x$ direction. Every function of the form $g(x + 2y)$ satisfies the same relation, which is how you recognise a travelling wave from its partial derivatives alone.