Mathematics RU

Practice · Chapter 32

Partial derivatives and the gradient

Partial derivatives of functions of two variables, the gradient at a point, the equation of the tangent plane and stationary points.

How to solve it

The partial derivative with respect to $x$ is an ordinary derivative in which $y$ is treated as a constant. The two partial derivatives together form the gradient, a vector pointing in the direction of the fastest growth.

Step by step

  1. $\frac{\partial f}{\partial x}$: differentiate in $x$, treating $y$ as a number. Terms without $x$ give zero.
  2. The gradient at a point: find both partial derivatives and substitute the coordinates.
  3. The tangent plane at $(x_0,\ y_0)$: $z = f(x_0, y_0) + f_x(x - x_0) + f_y(y - y_0)$.
  4. A stationary point: both partial derivatives are zero — solve the system.
The partial derivative in $x$ at the point. The partial derivative in $y$ at the point. The value of the function at the point of contact. Example: $f = 3x^2 + 4xy + 5$; $f_x = 6x + 4y$, $f_y = 4x$; at $(-2,\ 2)$ the gradient is $(-4,\ -8)$.

Common mistakes

  • Differentiating $y$ as a variable when taking $\frac{\partial}{\partial x}$: the term $y^2$ gives $0$, and $4xy$ gives $4y$.
  • Forgetting the chain rule: $\frac{\partial}{\partial x}\sin(2x - y) = 2\cos(2x - y)$.
  • Putting expressions instead of numbers into the tangent plane: substitute the point first.
  • Setting only one partial derivative to zero when looking for a stationary point.

Example