Second partials of sin(x + y^2)

One inner function, two different inner derivatives

The inside is $u = x + y^{2}$. Its derivative is $1$ in the $x$ direction and $2y$ in the $y$ direction, and that single difference drives everything on this page.

Because the inner $x$ derivative is $1$, the $x$ direction behaves like a plain single variable sine wave: $f_{xx} = -\sin(x+y^{2}) = -f$. Differentiating $f_x$ in $y$ pulls out the inner derivative $2y$.

The $y$ second derivative is the only one that needs the product rule, since $f_y = 2y\cos(x+y^{2})$ has $y$ in the leading factor and inside the cosine.

The mistake: forgetting the term that comes from the 2y

Writing $f_{yy} = -4y^{2}\sin(x+y^{2})$ is the standard error. It applies the chain rule to the cosine and never differentiates the $2y$ standing in front. Check it along the $x$ axis, where $y = 0$: the true value is $2\cos x$, while the incomplete version gives $0$ for every $x$, which would wrongly say the surface has no curvature in the $y$ direction anywhere along that line.

The reverse error also happens, keeping only $2\cos(x+y^{2})$ and dropping the chain rule term. Both terms are real, and they compete: near $y = 0$ the first dominates, while for large $|y|$ the $4y^{2}$ factor takes over.

Nothing similar happens in the $x$ direction, because the inner derivative there is the constant $1$. That asymmetry is the point of this example: the same outer function can need a product rule in one variable and not in the other.

Relations between the three second partials

Because everything is built from the same inner function, the three answers are tied together.

• $f_{xx} = -f$, so the $x$ cross sections are ordinary sine waves.
• $f_{xy} = 2y\,f_{xx}$, the mixed partial is the pure $x$ curvature scaled by the inner derivative $2y$.
• $f_{yy} = 2f_{x} + 4y^{2}f_{xx}$, one term from differentiating the $2y$ and one from the chain rule.

Clairaut symmetry holds: differentiating $f_y = 2y\cos(x+y^{2})$ in $x$ leaves the constant $2y$ alone and turns the cosine into $-\sin$, giving $-2y\sin(x+y^{2})$, the same mixed partial found the other way.

The level curves are the parabolas $x + y^{2} = c$, so the whole surface is one sine wave bent along parabolic ridges. That is why the mixed partial carries a factor of $y$: the ridges tilt more steeply the further you go from the $x$ axis.