Multivariable calculus

Partial Derivatives of e^(xyz)

Q: What is the mixed partial with respect to x then y?

Differentiate yz\,e^{xyz} with respect to y using the product rule: the factor yz gives z, and the exponential gives yz \cdot xz\,e^{xyz}. Collecting terms, \frac{\partial^2 f}{\partial y\, \partial x} = z(1 + xyz)e^{xyz}.

Q: Can any partial derivative of this function be zero?

Yes, but only through the coefficient. Since e^{xyz} is never zero, \frac{\partial f}{\partial x} = 0 requires yz = 0, meaning y = 0 or z = 0.

For f(x,y,z) = e^(xyz), the partial derivative with respect to x is yz times e^(xyz), with respect to y is xz times e^(xyz), and with respect to z is xy times e^(xyz). The exponential reproduces itself and the chain rule multiplies it by the derivative of the exponent xyz.

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

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

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

\frac{\partial f}{\partial z} = xy\,e^{xyz}

One chain rule, three inside derivatives

Write $f = e^u$ with $u = xyz$ . The outside function is the exponential, whose derivative is itself, so every partial derivative of $f$ carries the factor $e^{xyz}$ unchanged. All that changes from one variable to the next is $\frac{\partial u}{\partial x}$ , the derivative of the exponent.

Differentiating $u = xyz$ with respect to $x$ holds $y$ and $z$ fixed, so $u$ behaves like a constant $yz$ times $x$ , giving $yz$ . The same reasoning gives $xz$ and $xy$ for the other two.

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

Notice which letter is missing in each coefficient. The partial with respect to $x$ carries $yz$ , the two variables you held still. That is a reliable structural check: the variable you differentiated should not appear in the coefficient out front.

The mistake: differentiating the exponent and stopping

The most common wrong answer is $yz$ on its own, with the exponential dropped. The chain rule multiplies, it does not replace. The outside factor $e^{xyz}$ survives every differentiation because the exponential is its own derivative.

A second error is treating $e^{xyz}$ like a power and writing $xyz\,e^{xyz-1}$ . The power rule applies when the variable is in the base, and here the variable is in the exponent. Those are different rules and they never mix.

Wrong: $\frac{\partial f}{\partial x} = yz$ , which forgets the outside factor.
Wrong: $\frac{\partial f}{\partial x} = xyz\,e^{xyz-1}$ , which applies the power rule to an exponent.
Right: $\frac{\partial f}{\partial x} = yz\,e^{xyz}$ , outside derivative times inside derivative.

A symmetry worth exploiting

Multiply each partial derivative by its own variable and you get the same thing three times, namely $xyz\,e^{xyz}$ . So $x f_x = y f_y = z f_z$ , which is a quick consistency test on all three answers at once.

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

At $(1,1,1)$ all three partial derivatives equal $e$ , as the symmetry of the function demands. Any answer that fails to be symmetric under swapping the three letters is wrong before you check anything else.

Frequently asked questions

What is the mixed partial with respect to x then y?

Differentiate $yz\,e^{xyz}$ with respect to $y$ using the product rule: the factor $yz$ gives $z$ , and the exponential gives $yz \cdot xz\,e^{xyz}$ . Collecting terms, $\frac{\partial^2 f}{\partial y\, \partial x} = z(1 + xyz)e^{xyz}$ .

Can any partial derivative of this function be zero?

Yes, but only through the coefficient. Since $e^{xyz}$ is never zero, $\frac{\partial f}{\partial x} = 0$ requires $yz = 0$ , meaning $y = 0$ or $z = 0$ .