Multivariable calculus

Gradient of ln(1 + x^2 y^2) With Full Working

Q: Is the origin a minimum of \ln(1 + x^{2}y^{2})?

It is a minimum, but not an isolated one. The value f = 0 is achieved along both entire axes, so the origin sits on a valley floor rather than in a bowl. The second derivative test is inconclusive there, which is exactly what you should expect for a degenerate critical set.

For f(x, y) = ln(1 + x^2 y^2) the gradient is (2xy^2, 2x^2 y) divided by 1 + x^2 y^2. Differentiate the logarithm to get 1 over the inside, then multiply by the partial derivative of the inside. The gradient is the zero vector on both coordinate axes, where f reaches its minimum value 0.

f(x,y) = \ln\left(1 + x^{2}y^{2}\right)

\nabla f = \left\langle \frac{2xy^{2}}{1 + x^{2}y^{2}},\; \frac{2x^{2}y}{1 + x^{2}y^{2}} \right\rangle

\frac{\partial f}{\partial x} = \frac{2xy^{2}}{1 + x^{2}y^{2}}

\frac{\partial f}{\partial y} = \frac{2x^{2}y}{1 + x^{2}y^{2}}

One over the inside, times the derivative of the inside

Write $u = 1 + x^{2}y^{2}$ . Then $f = \ln u$ and $\frac{\partial f}{\partial x} = \frac{1}{u}\cdot\frac{\partial u}{\partial x}$ . The denominator is the same for both components, because it is just $u$ and $u$ does not care which variable you are differentiating.

\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\left(1 + x^{2}y^{2}\right) = 2xy^{2}, \qquad \frac{\partial u}{\partial y} = 2x^{2}y

\nabla f = \frac{1}{1 + x^{2}y^{2}}\left\langle 2xy^{2},\; 2x^{2}y \right\rangle = \frac{2xy}{1 + x^{2}y^{2}}\left\langle y,\; x \right\rangle

The constant $1$ inside the logarithm contributes nothing to either numerator, since the derivative of a constant is zero. It matters only because it keeps $u \geq 1$ , so the logarithm is defined everywhere in the plane and the denominator never hits zero.

Where the surface is flat, and how fast it climbs

Both components carry a factor of $xy$ . That makes the gradient the zero vector along the entire $x$ axis and the entire $y$ axis, not just at the origin.

\nabla f(x, 0) = \left\langle 0, 0 \right\rangle, \qquad \nabla f(0, y) = \left\langle 0, 0 \right\rangle

That fits the surface. On the axes $x^{2}y^{2} = 0$ and $f = \ln 1 = 0$ , which is the smallest value $f$ ever takes, so the axes form a valley floor of minima.

At $(1,1)$ the numerators are $2$ and $2$ , and the denominator is $2$ .

\nabla f(1,1) = \left\langle 1,\; 1 \right\rangle

Far from the axes the growth stalls. With $x = y = t$ large, each component behaves like $2t^{3}/t^{4} = 2/t$ , so the surface flattens out. That is logarithmic growth showing itself in the gradient.

The mistake: stopping at one over the inside

The most frequent wrong answer is $\frac{\partial f}{\partial x} = \frac{1}{1 + x^{2}y^{2}}$ , which uses the outer derivative and forgets the inner one. A second version keeps the inner derivative but attaches it to the wrong variable, giving $\frac{2x^{2}y}{1 + x^{2}y^{2}}$ for the $x$ partial.

Two checks catch both.

Symmetry: $f$ is unchanged when you swap $x$ and $y$ , so the $x$ component must become the $y$ component under that swap. $\frac{2xy^{2}}{1+x^{2}y^{2}}$ and $\frac{2x^{2}y}{1+x^{2}y^{2}}$ do exactly that.
The axes: the true $x$ partial must vanish at $(1, 0)$ , since $f$ is identically $0$ along the whole $x$ axis. The dropped-chain-rule answer gives $1$ there instead of $0$ .

Frequently asked questions

Why do both components share the denominator $1 + x^{2}y^{2}$ ?

Because that denominator comes from the outer function. For $f = \ln u$ every partial derivative is $\frac{1}{u}$ times a partial of $u$ , and $\frac{1}{u}$ does not depend on which variable you differentiate. Only the numerators differ.

Is the origin a minimum of $\ln(1 + x^{2}y^{2})$ ?

It is a minimum, but not an isolated one. The value $f = 0$ is achieved along both entire axes, so the origin sits on a valley floor rather than in a bowl. The second derivative test is inconclusive there, which is exactly what you should expect for a degenerate critical set.