Multivariable calculus

Partial Derivatives of x^2y + y^2z + z^2x

Q: How do I evaluate the gradient at a specific point?

Differentiate first, substitute second. At (1,2,3) the gradient is \langle 13, 13, 10 \rangle. If you substitute the numbers into f before differentiating, you turn f into a constant and every derivative collapses to zero.

For f(x,y,z) = x^2 y + y^2 z + z^2 x, the partial derivative with respect to x is 2xy + z^2, with respect to y is x^2 + 2yz, and with respect to z is y^2 + 2zx. Each one treats the other two variables as constants, so the single term that lacks that variable contributes nothing.

f(x,y,z) = x^2 y + y^2 z + z^2 x

\frac{\partial f}{\partial x} = 2xy + z^2

\frac{\partial f}{\partial y} = x^2 + 2yz

\frac{\partial f}{\partial z} = y^2 + 2zx

Differentiate one variable at a time

A partial derivative in three variables is still an ordinary one variable derivative. To find $\frac{\partial f}{\partial x}$ , freeze $y$ and $z$ at whatever values they hold and differentiate what is left as a function of $x$ alone.

Go term by term. In $x^2 y$ the factor $y$ is just a constant multiplier, so that term gives $2xy$ . The term $y^2 z$ contains no $x$ at all, so it gives $0$ . In $z^2 x$ the factor $z^2$ is constant, and the derivative of $x$ is $1$ , so that term gives $z^2$ .

\frac{\partial f}{\partial x} = 2xy + 0 + z^2 = 2xy + z^2

The function is cyclic: swapping $x \to y \to z \to x$ returns the same expression. That means you can rotate the answer instead of redoing the work, which is a genuine shortcut and also a way to catch an arithmetic slip.

\frac{\partial f}{\partial y} = x^2 + 2yz, \qquad \frac{\partial f}{\partial z} = y^2 + 2zx

The mistake: treating a constant as a zero

Students often decide that $y^2 z$ vanishes when differentiating with respect to $y$ , on the grounds that $z$ is being held constant. Constants multiply, they do not delete. Read the term as $z \cdot y^2$ , a constant times $y^2$ , and its $y$ derivative is $2yz$ .

The mirror image of that error is expecting every partial derivative to have three terms because $f$ has three terms. Here exactly one term is free of each variable, so each partial derivative has two terms. Count the surviving terms deliberately rather than by habit.

A term containing the variable: differentiate it, keeping the other letters as constant factors.
A term not containing the variable: it contributes $0$ .
A constant factor never disappears, it multiplies the derivative.

A check that catches a wrong coefficient

Every term of $f$ has total degree three, so $f$ is homogeneous of degree three and Euler's relation holds. It ties all three partial derivatives to the original function, so one substitution audits the whole answer.

x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} + z\frac{\partial f}{\partial z} = 3f

Test it at $(1,2,3)$ . There $f = 23$ , and the partials come out to $13$ , $13$ and $10$ . The left side is $1(13) + 2(13) + 3(10) = 69$ , and the right side is $3 \times 23 = 69$ . Had you written $xy$ instead of $2xy$ , the two sides would disagree and you would know at once.

Frequently asked questions

How do I evaluate the gradient at a specific point?

Differentiate first, substitute second. At $(1,2,3)$ the gradient is $\langle 13, 13, 10 \rangle$ . If you substitute the numbers into $f$ before differentiating, you turn $f$ into a constant and every derivative collapses to zero.

Do the mixed partials come out the same in either order?

Yes. A polynomial is smooth everywhere, so Clairaut's theorem applies. Differentiating $2xy + z^2$ with respect to $y$ gives $2x$ , and differentiating $x^2 + 2yz$ with respect to $x$ also gives $2x$ .