Multivariable calculus

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

The partials of 1/(1 + x^2 + y^2) are f_x = -2x/(1 + x^2 + y^2)^2 and f_y = -2y/(1 + x^2 + y^2)^2. Rewrite the function as (1 + x^2 + y^2) to the power -1: the power rule gives minus that expression to the power -2, and the inner partial supplies 2x or 2y.

f(x,y)=11+x2+y2f(x,y) = \frac{1}{1 + x^2 + y^2}
∂f∂x=−2x(1+x2+y2)2\frac{\partial f}{\partial x} = \frac{-2x}{(1 + x^2 + y^2)^2}
∂f∂y=−2y(1+x2+y2)2\frac{\partial f}{\partial y} = \frac{-2y}{(1 + x^2 + y^2)^2}

A negative power beats the quotient rule

The quotient rule works, but with a constant numerator the negative power form is faster and has fewer places to slip. Write f=u−1f = u^{-1} with u=1+x2+y2u = 1 + x^2 + y^2, then apply the power rule and the chain rule in the usual order.

∂f∂x=−u−2⋅∂u∂x=−2x(1+x2+y2)2\frac{\partial f}{\partial x} = -u^{-2}\cdot\frac{\partial u}{\partial x} = \frac{-2x}{(1 + x^2 + y^2)^2}
∂f∂y=−2y(1+x2+y2)2\frac{\partial f}{\partial y} = \frac{-2y}{(1 + x^2 + y^2)^2}

The exponent goes from −1-1 to −2-2, not to 00. Subtracting 11 from −1-1 moves away from zero, and that single step is where most of the sign and power errors on this function begin.

The squared denominator and the missing minus

Two things go wrong. The denominator must be squared, so −2x1+x2+y2\frac{-2x}{1+x^2+y^2} is off by a whole factor of uu. And the sign must be negative, because the surface falls away from the origin in every direction.

Compare all three versions at (1,1)(1,1), where u=3u = 3. The true fx=−29≈−0.222f_x = -\frac{2}{9} \approx -0.222. Losing the square gives −23≈−0.667-\frac{2}{3} \approx -0.667, and losing the minus gives +0.222+0.222. Neither matches a bump that peaks at height 11 over the origin and decreases outward.

How this compares with the Gaussian bump

This surface has the same peak height and the same radial symmetry as e−x2−y2e^{-x^2-y^2}, but it decays far more slowly. The partials show it directly: along the xx axis, ∣fx∣=2∣x∣(1+x2)2|f_x| = \frac{2|x|}{(1+x^2)^2} behaves like 2∣x∣3\frac{2}{|x|^3} for large ∣x∣|x|, an algebraic decay, while the Gaussian's partials die faster than any power of xx.

That difference is why 11+x2+y2\frac{1}{1+x^2+y^2} has a finite maximum slope but a long tail, and it is worth knowing which of the two a problem hands you before estimating anything far from the origin.

Frequently asked questions

Is the quotient rule wrong here?

No, it gives the same answer. With g=1g = 1 and h=1+x2+y2h = 1 + x^2 + y^2, the quotient rule yields 0⋅h−1⋅2xh2=−2xh2\frac{0\cdot h - 1\cdot 2x}{h^2} = \frac{-2x}{h^2}. The negative power route simply has fewer terms to keep track of.

What is the gradient at the origin?

The zero vector. Both partials carry a factor of xx or yy, so both vanish at (0,0)(0,0). Since the denominator is smallest there, that point is the global maximum with f=1f = 1.