Multivariable calculus

Second Partial Derivatives of x^3 y^2

For f(x,y) = x^3 y^2 the second partial derivatives are f_xx = 6xy^2, f_yy = 2x^3, and the mixed partial f_xy = 6x^2 y. Differentiating in the other order gives f_yx = 6x^2 y as well, which is what Clairaut's theorem predicts: every partial here is a polynomial, so all of them are continuous.

f(x,y)=x3y2f(x,y) = x^{3}y^{2}
∂f∂x=3x2y2\frac{\partial f}{\partial x} = 3x^{2}y^{2}
∂f∂y=2x3y\frac{\partial f}{\partial y} = 2x^{3}y

Second and mixed partials

∂2f∂x2=6xy2\frac{\partial^{2} f}{\partial x^{2}} = 6xy^{2}
∂2f∂y ∂x=6x2y\frac{\partial^{2} f}{\partial y\,\partial x} = 6x^{2}y
∂2f∂y2=2x3\frac{\partial^{2} f}{\partial y^{2}} = 2x^{3}

Taking the second partials one step at a time

A second partial is just a first partial done twice, so get the two first partials right before anything else. To differentiate x3y2x^{3}y^{2} with respect to xx, treat yy as a fixed number: the factor y2y^{2} is a constant multiplier and only x3x^{3} moves.

∂f∂x=3x2y2,∂f∂y=2x3y\frac{\partial f}{\partial x} = 3x^{2}y^{2}, \qquad \frac{\partial f}{\partial y} = 2x^{3}y

Now differentiate each of those again. Sending ∂f∂x\frac{\partial f}{\partial x} back through the xx derivative gives fxxf_{xx}, and sending ∂f∂y\frac{\partial f}{\partial y} through the yy derivative gives fyyf_{yy}.

∂2f∂x2=6xy2,∂2f∂y2=2x3\frac{\partial^{2} f}{\partial x^{2}} = 6xy^{2}, \qquad \frac{\partial^{2} f}{\partial y^{2}} = 2x^{3}

The mixed partial takes one derivative in each variable. Starting from 3x2y23x^{2}y^{2} and differentiating in yy gives 6x2y6x^{2}y.

∂2f∂y ∂x=∂∂y(3x2y2)=6x2y\frac{\partial^{2} f}{\partial y\,\partial x} = \frac{\partial}{\partial y}\left(3x^{2}y^{2}\right) = 6x^{2}y

Clairaut's theorem: the order does not matter here

Run the mixed partial the other way as a check. Start from ∂f∂y=2x3y\frac{\partial f}{\partial y} = 2x^{3}y and differentiate in xx.

∂2f∂x ∂y=∂∂x(2x3y)=6x2y\frac{\partial^{2} f}{\partial x\,\partial y} = \frac{\partial}{\partial x}\left(2x^{3}y\right) = 6x^{2}y

Same answer. Clairaut's theorem says this always happens as long as both mixed partials are continuous near the point, and for a polynomial every derivative of every order is a polynomial, so continuity is automatic on the whole plane.

That gives you a free error check on every homework problem: compute the mixed partial in both orders, and if the two disagree, one of your first partials is wrong.

The mistake students make

The frequent slip is differentiating the wrong variable at the second step because the notation is read backwards. In ∂2f∂y ∂x\frac{\partial^{2} f}{\partial y\,\partial x} the operator closest to ff acts first, so you differentiate in xx first and then in yy.

  • fxyf_{xy} in subscript notation means xx first, then yy, left to right.
  • ∂2f∂y ∂x\frac{\partial^{2} f}{\partial y\,\partial x} in fraction notation means the same thing, right to left.
  • For this function both orders give 6x2y6x^{2}y, so the reading order does not change the answer, but for the notation on an exam it matters.

The other slip is squaring the first derivative. fxxf_{xx} is not (fx)2\left(f_x\right)^{2}. Here (3x2y2)2=9x4y4\left(3x^{2}y^{2}\right)^{2} = 9x^{4}y^{4}, which is nothing like the correct 6xy26xy^{2}.

Frequently asked questions

Do I have to compute f_yx separately from f_xy?

Not for a function like this one. Since x3y2x^{3}y^{2} is a polynomial, its mixed partials are continuous everywhere and Clairaut's theorem guarantees fxy=fyxf_{xy} = f_{yx}. Computing both anyway is still a good way to catch an arithmetic error.

Why is f_yy free of y?

Because ff is only quadratic in yy. Differentiating y2y^{2} twice gives the constant 2, so fyy=2x3f_{yy} = 2x^{3} depends on xx alone. A third derivative in yy would be zero.