Multivariable calculus

Partial Derivatives of xyz/(1 + x^2 y^2 z^2)

Q: Why do all three partials look identical apart from one factor?

Because f depends on the variables only through u = xyz. The chain rule gives \frac{\partial f}{\partial x} = f'(u)\frac{\partial u}{\partial x}, and f'(u) is the same scalar for all three. Only \frac{\partial u}{\partial x} = yz changes as you move from one variable to the next.

Q: What is the largest value this function takes?

Since f = \frac{u}{1+u^{2}} with u = xyz, the maximum is \frac{1}{2}, reached whenever u = 1, for instance at (1,1,1) and at \left(2,1,\tfrac{1}{2}\right). The minimum is -\frac{1}{2} where u = -1. Those are exactly the surfaces on which the gradient vanishes, which is what you expect at an extreme value.

For f(x,y,z) = xyz/(1 + x^2 y^2 z^2), the partial with respect to x is yz(1 - x^2 y^2 z^2) divided by (1 + x^2 y^2 z^2) squared. The y and z partials replace the leading yz by xz and by xy. The function depends on all three variables only through the product xyz.

f(x,y,z) = \frac{xyz}{1 + x^{2}y^{2}z^{2}}

\frac{\partial f}{\partial x} = \frac{yz\left(1 - x^{2}y^{2}z^{2}\right)}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}

\frac{\partial f}{\partial y} = \frac{xz\left(1 - x^{2}y^{2}z^{2}\right)}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}

\frac{\partial f}{\partial z} = \frac{xy\left(1 - x^{2}y^{2}z^{2}\right)}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}

Everything runs through the single product xyz

Set $u = xyz$ . The whole function is $\frac{u}{1+u^{2}}$ , a one-variable function in a three-variable disguise. Differentiate once with respect to $u$ , then attach the partial of $u$ , and all three answers come out together.

\frac{d}{du}\left(\frac{u}{1+u^{2}}\right) = \frac{\left(1+u^{2}\right) - u(2u)}{\left(1+u^{2}\right)^{2}} = \frac{1-u^{2}}{\left(1+u^{2}\right)^{2}}

The partials of $u$ are the easy half: $\frac{\partial u}{\partial x} = yz$ , $\frac{\partial u}{\partial y} = xz$ , and $\frac{\partial u}{\partial z} = xy$ . Each one multiplies the derivative above.

\frac{\partial f}{\partial x} = \frac{yz\left(1 - x^{2}y^{2}z^{2}\right)}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}

\frac{\partial f}{\partial y} = \frac{xz\left(1 - x^{2}y^{2}z^{2}\right)}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}, \qquad \frac{\partial f}{\partial z} = \frac{xy\left(1 - x^{2}y^{2}z^{2}\right)}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}

The direct quotient rule agrees, as it must. In $x$ it gives $\frac{yz\,D - xyz\left(2xy^{2}z^{2}\right)}{D^{2}}$ with $D = 1+x^{2}y^{2}z^{2}$ , and pulling $yz$ out of the top leaves the bracket $D - 2x^{2}y^{2}z^{2} = 1 - x^{2}y^{2}z^{2}$ .

The mistake: differentiating the denominator as if only x were in it

The denominator is $1 + x^{2}y^{2}z^{2}$ , and its $x$ -derivative is $2xy^{2}z^{2}$ , not $2x$ . Losing the $y^{2}z^{2}$ is the standard error here, and it is a quiet one: at $y = z = 1$ the wrong derivative and the right one agree, so a check at $(x,1,1)$ will not expose it.

$\frac{\partial}{\partial x}\left(x^{2}y^{2}z^{2}\right) = 2xy^{2}z^{2}$ , since $y^{2}z^{2}$ is a constant multiplier when $x$ moves.
The leading factor is the derivative of the numerator, so it is $yz$ in the $x$ -partial, never $xyz$ .
The denominator is at least 1 everywhere, so no partial has a domain restriction.

Test a candidate answer at a point where the three variables differ, say $\left(2,1,\tfrac{1}{2}\right)$ rather than $(1,1,1)$ . Symmetric test points hide exactly the errors that confuse one variable with another.

The gradient vanishes on whole surfaces

All three partials share the factor $1-u^{2}$ , so the gradient collapses into one scalar times a simple vector.

\nabla f = \frac{1 - x^{2}y^{2}z^{2}}{\left(1 + x^{2}y^{2}z^{2}\right)^{2}}\,\langle yz,\; xz,\; xy \rangle

That shared factor is zero when $xyz = 1$ or $xyz = -1$ , so the gradient dies on two entire surfaces rather than at isolated points. The only other way to kill it is to make $yz$ , $xz$ and $xy$ vanish together, which needs at least two of the three variables to be zero, giving the three coordinate axes.

At $(1,1,2)$ the product is $u = 2$ , so $D = 5$ and the shared factor is $\frac{1-4}{25} = -\frac{3}{25}$ . Multiplying it by $\langle yz, xz, xy \rangle = \langle 2, 2, 1 \rangle$ gives the gradient.

\nabla f(1,1,2) = \left\langle -\tfrac{6}{25},\; -\tfrac{6}{25},\; -\tfrac{3}{25} \right\rangle

Every component is negative because $u = 2$ sits past the peak of $\frac{u}{1+u^{2}}$ at $u = 1$ . Beyond that peak any nudge that increases the product $xyz$ decreases $f$ .

Frequently asked questions

Why do all three partials look identical apart from one factor?

Because $f$ depends on the variables only through $u = xyz$ . The chain rule gives $\frac{\partial f}{\partial x} = f'(u)\frac{\partial u}{\partial x}$ , and $f'(u)$ is the same scalar for all three. Only $\frac{\partial u}{\partial x} = yz$ changes as you move from one variable to the next.

What is the largest value this function takes?

Since $f = \frac{u}{1+u^{2}}$ with $u = xyz$ , the maximum is $\frac{1}{2}$ , reached whenever $u = 1$ , for instance at $(1,1,1)$ and at $\left(2,1,\tfrac{1}{2}\right)$ . The minimum is $-\frac{1}{2}$ where $u = -1$ . Those are exactly the surfaces on which the gradient vanishes, which is what you expect at an extreme value.