Multivariable calculus

Gradient of x ln(1 + y^2) and Its Line of Zeros

Q: Why does the x component keep the logarithm?

Because nothing differentiates it. With respect to x, the factor \ln(1+y^{2}) is a number, and the derivative of a number times x is that number. The logarithm only turns into a rational function when you differentiate with respect to y.

Q: How large can the y component get?

For fixed x it is x times \frac{2y}{1+y^{2}}, and that factor has maximum value 1 at y = 1 and minimum -1 at y = -1. So the y component never exceeds \left|x\right| in absolute value, and the steepest climb in y always happens at y = \pm 1.

For f(x, y) = x ln(1 + y^2) the gradient is (ln(1 + y^2), 2xy over 1 + y^2). The x partial needs no work, since x appears to the first power with a constant multiplier. The y partial is a chain rule on the logarithm. Both components vanish along the whole x axis.

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

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

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

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

One easy component and one chain rule

For $\frac{\partial f}{\partial x}$ , the whole logarithm is a constant multiplier and $x$ is to the first power, so the derivative is the multiplier itself.

\frac{\partial f}{\partial x} = \ln\left(1+y^{2}\right)\cdot\frac{d}{dx}(x) = \ln\left(1+y^{2}\right)

The other component needs the chain rule inside the logarithm, with $x$ riding along as a constant.

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

Note the difference in flavour. One component keeps the logarithm and the other loses it entirely, because differentiating a logarithm produces a rational function. If both of your components contain a logarithm, or neither does, something has gone wrong.

A whole line of critical points

Set both components to zero. The first needs $\ln(1+y^{2}) = 0$ , which happens only at $y = 0$ . The second needs $2xy = 0$ , which is automatic once $y = 0$ .

\nabla f(x, 0) = \left\langle \ln 1,\; 0 \right\rangle = \left\langle 0,\; 0 \right\rangle \quad \text{for every } x

So the entire $x$ axis is critical. It is not a ridge of maxima or a trough of minima though. Since $\ln(1+y^{2}) > 0$ for $y \neq 0$ , the sign of $f$ near the axis is the sign of $x$ , so at a point like $(2,0)$ the surface rises on both sides in $y$ , while at $(-2, 0)$ it falls on both sides. The second derivative test is inconclusive along the whole line, and you have to look at the sign of $f$ instead.

Away from the axis, evaluate at $(2,1)$ : the logarithm gives $\ln 2 \approx 0.6931$ and the second component is $\frac{2\cdot 2\cdot 1}{2} = 2$ .

\nabla f(2,1) = \left\langle \ln 2,\; 2 \right\rangle \approx \left\langle 0.6931,\; 2 \right\rangle

The surface is climbing about three times faster in $y$ than in $x$ there, which is the kind of comparison a gradient is for.

The mistake: dropping the denominator, or the x

Two errors account for nearly all wrong answers on this one.

Writing $\frac{\partial f}{\partial y} = 2xy$ , which keeps the inner derivative and forgets the $\frac{1}{1+y^{2}}$ from the logarithm. That answer grows without bound in $y$ , while the true one is bounded, since $\frac{2y}{1+y^{2}}$ never exceeds $1$ in absolute value.
Writing $\frac{\partial f}{\partial y} = \frac{2y}{1+y^{2}}$ , dropping the factor $x$ . Then the surface would be predicted to change in $y$ even along the line $x = 0$ , where $f$ is identically zero.

There is also a domain point worth stating: $1 + y^{2} \geq 1$ for every real $y$ , so the logarithm is defined on the whole plane and the denominator is never zero. That is why this function is a safe example, unlike $x\ln y$ , which needs $y > 0$ .

Frequently asked questions

Why does the $x$ component keep the logarithm?

Because nothing differentiates it. With respect to $x$ , the factor $\ln(1+y^{2})$ is a number, and the derivative of a number times $x$ is that number. The logarithm only turns into a rational function when you differentiate with respect to $y$ .

How large can the $y$ component get?

For fixed $x$ it is $x$ times $\frac{2y}{1+y^{2}}$ , and that factor has maximum value $1$ at $y = 1$ and minimum $-1$ at $y = -1$ . So the $y$ component never exceeds $\left|x\right|$ in absolute value, and the steepest climb in $y$ always happens at $y = \pm 1$ .