Multivariable calculus

Partial Derivatives of xyz e^z

Q: Why does the z-partial have a (z + 1) factor?

The product rule on z\,e^{z} gives e^{z} + z\,e^{z}. Both terms share e^{z}, so factoring it out leaves 1 + z. That factored form is more useful than the expanded one, because it shows immediately where the derivative is zero.

Q: Do the mixed partials commute for this function?

Yes. For example, differentiating \frac{\partial f}{\partial x} = yz\,e^{z} with respect to z gives y(z+1)e^{z}, and differentiating \frac{\partial f}{\partial z} = xy(z+1)e^{z} with respect to x gives y(z+1)e^{z} as well. The function is smooth on all of space, so Clairaut's theorem applies.

For f(x,y,z) = xyz e^z, the partial with respect to x is yz e^z, the partial with respect to y is xz e^z, and the partial with respect to z is xy(z + 1)e^z. Only the z-partial needs the product rule, since z is the only variable appearing in two factors at once.

f(x,y,z) = xyz\,e^{z}

\frac{\partial f}{\partial x} = yz\,e^{z}

\frac{\partial f}{\partial y} = xz\,e^{z}

\frac{\partial f}{\partial z} = xy(z+1)e^{z}

Find the variable that appears twice

With three variables the bookkeeping matters more, so start by asking which variable shows up in more than one factor. Here $x$ appears once, $y$ appears once, and $z$ appears twice: as the bare factor $z$ and inside $e^{z}$ . Only $z$ can trigger the product rule.

For the $x$ -partial, everything except $x$ is frozen, so $f$ is the constant $yz\,e^{z}$ times $x$ . The same reasoning gives the $y$ -partial.

\frac{\partial f}{\partial x} = yz\,e^{z}, \qquad \frac{\partial f}{\partial y} = xz\,e^{z}

For the $z$ -partial, freeze $x$ and $y$ so the constant $xy$ multiplies the genuine product $z\,e^{z}$ . Now apply the product rule and factor the common exponential.

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

The mistake: assuming every variable behaves the same way

In a symmetric looking expression like $xyz$ , it is tempting to write all three partials by the same pattern and produce $xy\,e^{z}$ for the $z$ -partial. That answer misses the term where the exponential is differentiated. It happens to agree with the truth where $z = 0$ , and also wherever $xy = 0$ makes both versions vanish, but it is wrong everywhere else.

Only $z$ appears in two $z$ -dependent factors, so only the $z$ -partial gets a product rule.
The factored form $xy(z+1)e^{z}$ shows the $z$ -partial vanishes when $z = -1$ , or when $x = 0$ , or when $y = 0$ .
The exponential never contributes a zero, since $e^{z} > 0$ for every real $z$ .

A structural check: the $x$ -partial should contain no $x$ , and the $y$ -partial should contain no $y$ , because $f$ is degree one in each of those variables. The $z$ -partial does still contain $z$ , which is the visible fingerprint of the product rule.

The gradient at a point on the plane z = 0

Take $(1,1,0)$ . Both the $x$ -partial and the $y$ -partial carry a factor of $z$ , so they vanish. The $z$ -partial does not, because of the $+1$ produced by the product rule.

\nabla f(1,1,0) = \langle 0,\; 0,\; 1 \rangle

Near $(1,1,0)$ the set where $f$ vanishes is exactly the plane $z=0$ , since $x$ and $y$ are nowhere near zero there. A gradient is perpendicular to its level set, so $\langle 0,0,1\rangle$ is the only direction it could point. That agreement is a strong check on the algebra.

Frequently asked questions

Why does the z-partial have a (z + 1) factor?

The product rule on $z\,e^{z}$ gives $e^{z} + z\,e^{z}$ . Both terms share $e^{z}$ , so factoring it out leaves $1 + z$ . That factored form is more useful than the expanded one, because it shows immediately where the derivative is zero.

Do the mixed partials commute for this function?

Yes. For example, differentiating $\frac{\partial f}{\partial x} = yz\,e^{z}$ with respect to $z$ gives $y(z+1)e^{z}$ , and differentiating $\frac{\partial f}{\partial z} = xy(z+1)e^{z}$ with respect to $x$ gives $y(z+1)e^{z}$ as well. The function is smooth on all of space, so Clairaut's theorem applies.