Multivariable calculus

Second Partial Derivatives of e^(xy)

Q: Why does the mixed partial of e^(xy) need a product rule when the pure ones do not?

Because f_x = y\,e^{xy} has a loose factor of y out front. Differentiating again in x leaves that factor alone, but differentiating in y hits both it and the exponential, which is where the extra e^{xy} term comes from.

Q: Is f_xy = f_yx here?

Yes. Both equal (1 + xy)e^{xy}. Every partial of e^{xy} is a polynomial times e^{xy}, so all of them are continuous on the whole plane and Clairaut's theorem applies at every point.

For f(x,y) = e^(xy) the second partials are f_xx = y^2 e^(xy), f_yy = x^2 e^(xy), and f_xy = (1 + xy) e^(xy). The mixed partial needs the product rule, since f_x = y e^(xy) is a product of y and the exponential. Both orders give the same (1 + xy) e^(xy), as Clairaut's theorem requires.

f(x,y) = e^{xy}

\frac{\partial f}{\partial x} = y\,e^{xy}

\frac{\partial f}{\partial y} = x\,e^{xy}

Second and mixed partials

\frac{\partial^{2} f}{\partial x^{2}} = y^{2}e^{xy}

\frac{\partial^{2} f}{\partial y\,\partial x} = (1 + xy)\,e^{xy}

\frac{\partial^{2} f}{\partial y^{2}} = x^{2}e^{xy}

Chain rule first, then product rule

Differentiating $e^{xy}$ in $x$ is a chain rule: the outside is $e^{u}$ and the inside is $u = xy$ , whose $x$ derivative is $y$ .

\frac{\partial f}{\partial x} = y\,e^{xy}, \qquad \frac{\partial f}{\partial y} = x\,e^{xy}

For $f_{xx}$ , the factor $y$ is constant while $x$ moves, so you only chain through the exponential again and pick up a second factor of $y$ .

\frac{\partial^{2} f}{\partial x^{2}} = y \cdot y\,e^{xy} = y^{2}e^{xy}, \qquad \frac{\partial^{2} f}{\partial y^{2}} = x^{2}e^{xy}

The mixed partial is different in character. Differentiate $f_x = y\,e^{xy}$ with respect to $y$ and both factors depend on $y$ , so the product rule is unavoidable.

\frac{\partial}{\partial y}\left(y\,e^{xy}\right) = 1 \cdot e^{xy} + y \cdot x\,e^{xy} = (1 + xy)\,e^{xy}

The mistake students make

Almost every wrong answer here is $f_{xy} = xy\,e^{xy}$ , which drops the $e^{xy}$ term that comes from differentiating the loose factor of $y$ . The tell is the value at the origin: the correct mixed partial gives $f_{xy}(0,0) = 1$ , while the dropped term version gives $0$ .

Check it from the other side. Differentiate $f_y = x\,e^{xy}$ with respect to $x$ and the product rule gives $e^{xy} + x \cdot y\,e^{xy}$ , the same $(1 + xy)e^{xy}$ . Two independent routes to one answer is exactly the Clairaut check, and it catches the missing term immediately.

\frac{\partial^{2} f}{\partial x\,\partial y} = e^{xy} + xy\,e^{xy} = (1 + xy)\,e^{xy} = \frac{\partial^{2} f}{\partial y\,\partial x}

A second slip is writing $f_{xx} = ye^{xy}$ , forgetting that the chain rule contributes another $y$ each time you differentiate in $x$ . Every extra $x$ derivative multiplies by $y$ , so the $n$ th one is $y^{n}e^{xy}$ .

Reading the Hessian

Collect the three second partials into the Hessian matrix, the object the second derivative test uses.

H = e^{xy}\begin{pmatrix} y^{2} & 1 + xy \\ 1 + xy & x^{2} \end{pmatrix}

Its determinant is $e^{2xy}\left(x^{2}y^{2} - (1+xy)^{2}\right) = -e^{2xy}(1 + 2xy)$ . Clairaut symmetry is what makes this matrix symmetric, and symmetry is why the second derivative test only needs three numbers rather than four.

Frequently asked questions

Why does the mixed partial of e^(xy) need a product rule when the pure ones do not?

Because $f_x = y\,e^{xy}$ has a loose factor of $y$ out front. Differentiating again in $x$ leaves that factor alone, but differentiating in $y$ hits both it and the exponential, which is where the extra $e^{xy}$ term comes from.

Is f_xy = f_yx here?

Yes. Both equal $(1 + xy)e^{xy}$ . Every partial of $e^{xy}$ is a polynomial times $e^{xy}$ , so all of them are continuous on the whole plane and Clairaut's theorem applies at every point.