Multivariable calculus

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

Q: Why do the x and y partial derivatives come out equal?

Because the function depends on x and y only through the sum x + y, and that sum changes at the same rate whichever of the two you move. Any function of x+y has equal partial derivatives in x and y.

Q: Is this function defined everywhere?

Yes. The denominator 1 + z^2 is at least 1 for every real z, so it never vanishes and no point has to be excluded. That is what makes the quotient safe to differentiate anywhere.

For f(x,y,z) = (x + y)/(1 + z^2), the partial derivatives with respect to x and with respect to y are both 1/(1 + z^2), and the partial with respect to z is negative 2z(x + y) divided by (1 + z^2) squared. Only the z partial needs the quotient rule.

f(x,y,z) = \frac{x + y}{1 + z^2}

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

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

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

Ask which variable is actually in the denominator

A fraction does not automatically call for the quotient rule. What matters is whether the differentiating variable appears in the denominator. Here $x$ and $y$ appear only upstairs, so for those two the whole factor $\frac{1}{1+z^2}$ is a constant multiplier.

That makes the first two partial derivatives almost immediate. Differentiating $x + y$ with respect to $x$ gives $1$ , and the constant factor comes along.

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

For $z$ the situation is different, since $z$ sits in the denominator. Treat $x + y$ as a constant $C$ and differentiate $\frac{C}{1+z^2}$ . The quotient rule gives numerator $0 \cdot (1+z^2) - C \cdot 2z$ over $(1+z^2)^2$ .

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

The mistake: reaching for the quotient rule every time

Running the quotient rule for $\frac{\partial f}{\partial x}$ is not wrong, it is just slower and it invites algebra errors. It gives $\frac{1 \cdot (1+z^2) - (x+y)\cdot 0}{(1+z^2)^2}$ , which simplifies back to $\frac{1}{1+z^2}$ . Students who do not simplify often leave an answer that looks different from the correct one and then distrust it.

The genuine error to avoid is a sign. Differentiating a function that sits in a denominator produces a minus sign, and it is easy to lose when you are also juggling the constant $x + y$ . Rewriting as $(x+y)(1+z^2)^{-1}$ makes the sign automatic: the power rule turns $-1$ into $-2$ and leaves a factor of $-1$ out front.

\frac{\partial}{\partial z}\left[(x+y)(1+z^2)^{-1}\right] = (x+y)\cdot(-1)(1+z^2)^{-2}\cdot 2z

Reading the answers back

At $(1,2,1)$ the denominator $1 + z^2$ is $2$ , so both the $x$ and $y$ partials equal $\frac{1}{2}$ . The $z$ partial is $\frac{-2(1)(3)}{4} = -\frac{3}{2}$ .

The signs are worth interpreting. Increasing $x$ or $y$ increases the numerator, so those partials are positive everywhere. The $z$ partial carries the factor $-2z(x+y)$ , so when $x + y > 0$ it is negative for $z > 0$ and positive for $z < 0$ : moving $z$ away from zero in either direction grows the denominator and shrinks the fraction. On the plane $z = 0$ the $z$ partial vanishes, since the denominator is momentarily flat.

Frequently asked questions

Why do the x and y partial derivatives come out equal?

Because the function depends on $x$ and $y$ only through the sum $x + y$ , and that sum changes at the same rate whichever of the two you move. Any function of $x+y$ has equal partial derivatives in $x$ and $y$ .

Is this function defined everywhere?

Yes. The denominator $1 + z^2$ is at least $1$ for every real $z$ , so it never vanishes and no point has to be excluded. That is what makes the quotient safe to differentiate anywhere.