Multivariable calculus

Partial Derivatives of cos(x^2 y): Chain Rule

Q: Why is there no y in the y partial?

Because u = x^2y is linear in y, so \frac{\partial u}{\partial y} = x^2 with no y remaining. The variable has not vanished from the problem though: it is still there inside \sin(x^2y).

The partials of cos(x^2 y) are f_x = -2xy sin(x^2 y) and f_y = -x^2 sin(x^2 y). The outer cosine becomes minus sine of the same inner expression each time. The inner partial of x^2 y is 2xy with respect to x, since y is a constant multiplier, and x^2 with respect to y.

f(x,y) = \cos(x^2 y)

\frac{\partial f}{\partial x} = -2xy\sin(x^2 y)

\frac{\partial f}{\partial y} = -x^2\sin(x^2 y)

The inner function is a product, but only one factor varies

Put $u = x^2y$ . Differentiating $u$ with respect to $x$ treats $y$ as a constant coefficient and gives $2xy$ . Differentiating with respect to $y$ treats $x^2$ as a constant coefficient and gives $x^2$ . Neither step needs the product rule, because in each case one factor is frozen.

\frac{\partial f}{\partial x} = -\sin(x^2 y)\cdot 2xy = -2xy\sin(x^2 y)

\frac{\partial f}{\partial y} = -\sin(x^2 y)\cdot x^2 = -x^2\sin(x^2 y)

Notice the outer factor $-\sin(x^2y)$ is identical in both. Only the inner partial distinguishes them, which is true of every chain rule problem with a single inner function.

Reaching for the product rule too early

Students often try to apply the product rule to $\cos(x^2y)$ itself, as if it were $\cos(x^2)\cdot y$ . It is not a product. It is a single cosine of a product, and the product lives entirely inside the argument where the constant-multiple rule handles it.

The other frequent error is losing the $y$ from the $x$ partial and writing $-2x\sin(x^2y)$ . At $(1, 3)$ that gives $-2\sin 3 \approx -0.282$ instead of the true $-6\sin 3 \approx -0.847$ , a factor of $3$ out, which is exactly the missing $y$ .

Identify the inner function first, in this case $x^2y$ .
Differentiate the outer cosine, keeping the inner expression intact.
Differentiate the inner function with respect to one variable only.

Where the surface is flat

Both partials vanish exactly where $\sin(x^2y) = 0$ , that is on the curves $x^2y = k\pi$ for integer $k$ . Nothing extra comes from the factors $x$ and $x^2$ out front: $f_y = -x^2\sin(x^2y)$ can only lose its $x^2$ at $x = 0$ , and there the argument $x^2y$ is $0$ , so the sine is zero anyway. The case $k = 0$ is worth reading slowly, because $x^2y = 0$ means $x = 0$ or $y = 0$ : both coordinate axes are already in the family.

Along the $y$ axis the function is the constant $\cos 0 = 1$ . That is a ridge at the maximum height of a cosine, not an isolated extremum, so the second derivative test cannot classify it and the bound $f \le 1$ has to.

Frequently asked questions

Does the chain rule apply more than once here?

Once. The outer cosine is the only composition, and ordinary single variable differentiation handles the inner $x^2y$ . You would chain again only if the inner function were itself a composition, which $x^2y$ is not.

Why is there no y in the y partial?

Because $u = x^2y$ is linear in $y$ , so $\frac{\partial u}{\partial y} = x^2$ with no $y$ remaining. The variable has not vanished from the problem though: it is still there inside $\sin(x^2y)$ .