Partial Derivatives

In multivariable calculus, we differentiate with respect to one variable while holding the others constant. In algebra competitions, this is primarily used for optimizing multivariable functions and Lagrange Multipliers.

Definition

Definition (Partial Derivative)

For a function

f (x_{1}, \dots, x_{n})

, the partial derivative with respect to

x_{i}

is denoted by

\frac{\partial f}{\partial x _{i}}

f_{x_{i}}

. It measures the rate of change of

f

in the direction of

x_{i}

Gradient and Optimization

The vector of all partial derivatives is the gradient, denoted $\nabla f$ .

\nabla f = (\frac{\partial f}{\partial x _{1}}, \dots, \frac{\partial f}{\partial x _{n}})

Critical Points

To find the local maxima or minima of a smooth function

f

, we set the gradient to zero:

\nabla f = 0 ⟹ \frac{\partial f}{\partial x _{i}} = 0 for all i

Example (Minimize Distance)

Find the point on the plane

x + 2 y + 3 z = 6

closest to the origin.

Proof. Minimize squared distance $f (x, y, z) = x^{2} + y^{2} + z^{2}$ subject to $z = (6 - x - 2 y) /3$ . Substitute $z$ and set partials $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ . (Alternatively, use Cauchy-Schwarz or Lagrange Multipliers). ∎

Lagrange Multipliers

To optimize

f (x, y)

subject to a constraint

g (x, y) = k

\nabla f = λ \nabla g

This creates a system of equations:

$f_{x} = λ g_{x}$
$f_{y} = λ g_{y}$
$g (x, y) = k$

Practice Problems

Exercise (Problem 1)

Find the minimum value of

x y + yz + z x

subject to

x + y + z = 1

using partial derivatives (substitute

z = 1 - x - y

Exercise (Problem 2)

Use Lagrange multipliers to maximize

x^{2} y

subject to

x + y = 3

Exercise (Problem 3)

Find the critical points of

f (x, y) = x^{3} + y^{3} - 3 x y