Partial Derivatives of ln(x^2 + y^2 + 1)

Use the inside derivative over the inside

For a logarithm of anything, $\frac{\partial}{\partial x}\ln(u) = \frac{1}{u}\frac{\partial u}{\partial x}$. Here $u = x^{2}+y^{2}+1$. Holding $y$ fixed makes both $y^{2}$ and $1$ constants, so $\frac{\partial u}{\partial x} = 2x$.

At $(1,2)$ the inside is $1 + 4 + 1 = 6$, so the partials are $\frac{2}{6} = \frac{1}{3}$ and $\frac{4}{6} = \frac{2}{3}$. The denominator is shared, which makes the ratio of the two slopes just the ratio $x : y$.

The mistake: splitting the logarithm of a sum

• $\ln(x^{2}+y^{2}+1)$ is not $\ln(x^{2}) + \ln(y^{2}) + \ln(1)$. Logarithms split across products and quotients, never across sums, so there is no shortcut past the chain rule here.
• Writing $\frac{\partial f}{\partial x} = \frac{1}{x^{2}+y^{2}+1}$ forgets the inside derivative $2x$. That version would be nonzero on the whole $y$-axis, where the true slope in $x$ is $0$.
• Writing $\frac{2x+2y}{x^{2}+y^{2}+1}$ differentiates the frozen variable as well.

Test any candidate at the origin. The surface has its lowest point there, $f(0,0) = 0$, so both partials must be $0$. The correct formulas give $\frac{0}{1} = 0$; the version missing the inside derivative gives $1$.

Domain and the radial picture

The argument $x^{2}+y^{2}+1$ is at least $1$, so it is never zero or negative and no domain restriction is needed. This is why courses use the plus 1 version: $\ln(x^{2}+y^{2})$ has partials of the same shape, $\frac{2x}{x^{2}+y^{2}}$ and $\frac{2y}{x^{2}+y^{2}}$, but both of those blow up at the origin, and $\ln(xy)$ is undefined on half the plane.

The gradient is a positive multiple of the position vector, so steepest ascent is straight away from the origin and the level curves are circles. The gradient length is $\frac{2r}{1+r^{2}}$, which is largest on the unit circle $r = 1$, where it equals exactly $1$.