Partial Derivatives of sqrt(1 + x^2 + 2y^2 + 3z^2)

One pattern, three different weights

Let $u = 1 + x^2 + 2y^2 + 3z^2$, so $f = \sqrt{u}$ and the outer derivative is $\frac{1}{2\sqrt{u}}$. The inner partials are $2x$, $4y$ and $6z$, and the $2$ in that denominator halves each of them.

Halving does not erase the weights, it passes them through. A coefficient $c$ on a squared term comes out as a coefficient $c$ on that variable, which is exactly why the numerators read $x$, $2y$ and $3z$.

Assuming every 2 cancels

On the unweighted version of this function all three coefficients cancel and the partials are $x$, $y$ and $z$ over the root. That memory is the trap. At $(1,1,1)$ the inner sum is $7$, so the partials are $\frac{1}{\sqrt 7} \approx 0.378$, $\frac{2}{\sqrt 7} \approx 0.756$ and $\frac{3}{\sqrt 7} \approx 1.134$: they sit in the ratio $1:2:3$ even though the three coordinates are equal.

A second slip is losing either the $\frac{1}{2}$ from the outer derivative or the inner partial itself, each of which puts the answer out by a factor of $2$. A third is over-simplifying the denominator: $\frac{x}{\sqrt{1 + x^2 + 2y^2 + 3z^2}}$ is not $\frac{x}{1 + x + 2y + 3z}$ and it is not $\frac{1}{\sqrt{1 + 2y^2 + 3z^2}}$. Square roots do not distribute across sums, so the denominator stays as one root.

• Differentiate the outer root first, leaving the whole sum inside it.
• Differentiate the inner sum with respect to one variable only.
• Multiply, halve, then check the coefficient actually survived.

How steep it can get

The gradient is $\frac{(x,\,2y,\,3z)}{\sqrt{u}}$, so its length squared is $\frac{x^2 + 4y^2 + 9z^2}{1 + x^2 + 2y^2 + 3z^2}$. Term by term the numerator carries the square of each weight while the denominator carries the weight itself, giving ratios $\frac{1}{1}$, $\frac{4}{2}$ and $\frac{9}{3}$. The largest is $3$, so $|\nabla f| < \sqrt{3} \approx 1.732$ everywhere.

That bound is approached but never reached. Run out along the $z$ axis and the quotient tends to $\frac{9z^2}{1 + 3z^2} \to 3$; do the same along the $x$ axis and it tends to $1$. The function is steepest in the direction it is weighted most heavily, which is what the coefficients were always going to say.

Every numerator carries its own variable, so all three partials vanish only at $(0,0,0)$. That is the sole critical point and the global minimum, with $f = 1$, because $u$ is at its smallest there and the square root is increasing.