Multivariable calculus

Partial Derivatives of arctan(x + yz)

Q: Where is the x partial derivative largest?

Wherever x + yz = 0, since that makes the denominator as small as possible, namely 1. There \frac{\partial f}{\partial x} = 1, which is the steepest the arctangent ever gets in the x direction.

Q: Can the y partial derivative be negative?

Yes. The denominator is always positive, so the sign of \frac{\partial f}{\partial y} is the sign of z. It is negative whenever z < 0 and zero on the plane z = 0.

For f(x,y,z) = arctan(x + yz), the partial derivative with respect to x is 1 over 1 plus (x + yz) squared. The y partial is z times that, and the z partial is y times that. The arctangent supplies one common denominator and the chain rule supplies the numerators 1, z and y.

f(x,y,z) = \arctan(x + yz)

\frac{\partial f}{\partial x} = \frac{1}{1 + (x + yz)^2}

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

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

One outside derivative, three inside derivatives

Let $u = x + yz$ , so $f = \arctan u$ . The derivative of $\arctan u$ is $\frac{1}{1+u^2}$ times the derivative of $u$ . Since $u$ is the same for all three variables, the denominator $1 + (x+yz)^2$ is identical in all three answers, and only the inside derivative changes.

Differentiate $u = x + yz$ one variable at a time. With respect to $x$ the $yz$ term is constant, so $\frac{\partial u}{\partial x} = 1$ . With respect to $y$ the $x$ is constant and $yz$ behaves like a constant $z$ times $y$ , so $\frac{\partial u}{\partial y} = z$ . Symmetrically $\frac{\partial u}{\partial z} = y$ .

\frac{\partial f}{\partial y} = \frac{1}{1 + u^2}\cdot z = \frac{z}{1 + (x + yz)^2}

So the three answers are the same fraction with numerators $1$ , $z$ and $y$ . Compute the denominator once and reuse it, which is faster and removes a chance to make three different transcription errors.

The mistake: squaring the wrong thing

The denominator is $1 + u^2$ where $u$ is the entire inside expression, so it is $1 + (x+yz)^2$ and not $1 + x^2 + y^2z^2$ . Those differ by the cross term $2xyz$ , which is not zero in general. Keep the parentheses in place through the whole calculation.

The second slip is swapping the numerators of the $y$ and $z$ partials. Differentiating with respect to $y$ leaves $z$ behind, not $y$ . The general rule holds: the variable you differentiated should not survive in the numerator that the chain rule produces.

Wrong denominator: $1 + x^2 + y^2z^2$ , which drops the cross term.
Wrong numerator: $y$ for the $y$ partial, when it should be $z$ .
Right: one denominator $1 + (x+yz)^2$ , numerators $1$ , $z$ and $y$ .

Structure and a sample point

Because the denominator is shared, the other two partial derivatives are simple multiples of the first: $f_y = z f_x$ and $f_z = y f_x$ . That relation is a fast internal consistency check on an answer you have just written.

At $(0,1,1)$ the inside is $x + yz = 1$ , so the denominator is $2$ and all three partial derivatives equal $\frac{1}{2}$ . Note also that the denominator is at least $1$ everywhere, so $f_x$ can never exceed $1$ in size. The other two are not bounded, because they carry the extra factors $z$ and $y$ : at $(0,0,10)$ the denominator is $1$ and $\frac{\partial f}{\partial y} = 10$ .

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

Frequently asked questions

Where is the x partial derivative largest?

Wherever $x + yz = 0$ , since that makes the denominator as small as possible, namely $1$ . There $\frac{\partial f}{\partial x} = 1$ , which is the steepest the arctangent ever gets in the $x$ direction.

Can the y partial derivative be negative?

Yes. The denominator is always positive, so the sign of $\frac{\partial f}{\partial y}$ is the sign of $z$ . It is negative whenever $z < 0$ and zero on the plane $z = 0$ .