Multivariable calculus

Partial Derivatives of y^2 cos(xy)

Q: Why is the x partial zero along the whole x axis?

On the line y = 0 the factor -y^3 is zero, so \frac{\partial f}{\partial x} = 0 there. That matches the function itself, which is identically zero along y = 0 and therefore flat in the x direction.

Q: Where do both partials vanish at once?

Setting y = 0 makes both zero, so the entire x axis consists of critical points. Away from that line, -y^3\sin(xy) = 0 forces \sin(xy) = 0, and substituting into the second partial leaves 2y\cos(xy) \neq 0.

For f(x, y) = y squared times cos(xy), the partial derivative with respect to x is negative y cubed times sin(xy), and the partial derivative with respect to y is 2y cos(xy) minus x y squared times sin(xy). The x partial needs only the chain rule; the y partial needs the product rule and the chain rule together.

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

\frac{\partial f}{\partial x} = -y^3\sin(xy)

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

The easy partial and the hard one

With $y$ fixed, the factor $y^2$ is a constant and only the cosine moves. The chain rule gives $-\sin(xy)$ times the inside derivative $y$ , and the leading $y^2$ multiplies through.

\frac{\partial f}{\partial x} = y^2\cdot\left(-\sin(xy)\right)\cdot y = -y^3\sin(xy)

With $x$ fixed, both factors contain $y$ , so the product rule is unavoidable. Differentiate $y^2$ to get $2y$ , keep $\cos(xy)$ , then keep $y^2$ and differentiate $\cos(xy)$ by $y$ , which brings out $-\sin(xy)$ times the inside derivative $x$ .

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

At $(0,2)$ the inside is zero, so $\frac{\partial f}{\partial x} = -8\sin 0 = 0$ and $\frac{\partial f}{\partial y} = 4\cos 0 - 0 = 4$ .

The mistake: using the same rule for both variables

Because the $y$ partial needs a product rule, students often add a spurious product rule term to the $x$ partial, tacking an extra piece onto $-y^3\sin(xy)$ . There is nothing to add: $y^2$ has zero derivative with respect to $x$ , so that branch of the product rule contributes zero.

The number of terms in a partial derivative depends on how many factors carry the moving variable, not on how the function looks overall.
Here $x$ appears in one factor, so $\frac{\partial f}{\partial x}$ has one term. $y$ appears in two factors, so $\frac{\partial f}{\partial y}$ has two.
Watch the exponent: $y^2$ times the inside derivative $y$ gives $y^3$ , not $y^2$ .

Checking with Clairaut's theorem

A good way to catch an error in either partial is to compute the mixed second derivative both ways and see whether they match. Starting from $\frac{\partial f}{\partial x} = -y^3\sin(xy)$ and differentiating by $y$ , the product rule gives $-3y^2\sin(xy) - xy^3\cos(xy)$ .

Starting instead from $\frac{\partial f}{\partial y} = 2y\cos(xy) - xy^2\sin(xy)$ and differentiating by $x$ , you get $-2y^2\sin(xy) - y^2\sin(xy) - xy^3\cos(xy)$ , and the first two terms combine to $-3y^2\sin(xy)$ .

\frac{\partial^2 f}{\partial y\,\partial x} = \frac{\partial^2 f}{\partial x\,\partial y} = -3y^2\sin(xy) - xy^3\cos(xy)

They agree, as they must for a smooth function. If your two routes had disagreed, one of the first partials would have been wrong.

Frequently asked questions

Why is the x partial zero along the whole x axis?

On the line $y = 0$ the factor $-y^3$ is zero, so $\frac{\partial f}{\partial x} = 0$ there. That matches the function itself, which is identically zero along $y = 0$ and therefore flat in the $x$ direction.

Where do both partials vanish at once?

Setting $y = 0$ makes both zero, so the entire $x$ axis consists of critical points. Away from that line, $-y^3\sin(xy) = 0$ forces $\sin(xy) = 0$ , and substituting into the second partial leaves $2y\cos(xy) \neq 0$ .