Directional Derivative Calculator

Published on: July 21, 2026
Final Answer: Free Full Steps: Pro

This Directional Derivative Calculator finds how fast \(f(x, y)\) changes in a chosen direction at a point. It computes the gradient, converts the direction into a unit vector, dots them together, and evaluates - showing every step.

Step-by-step method

  1. Set up the function, the direction vector, and the point.
  2. Find the gradient by taking the partial derivatives.
  3. Turn the direction into a unit vector.
  4. Dot the gradient with the unit vector.
  5. Evaluate at the given point.

Formula:

\(D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}, \quad \mathbf{u} = \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert}\)

Example 1:

\(f = x^{2} y,\; \mathbf{v} = \langle 3, 4 \rangle,\; P = (1, 2)\)

Step 1 - Set up the function, the direction vector, and the point.

In this problem: We are given \(f = x^{2} y\), direction \(\mathbf{v} = \langle 3, 4 \rangle\), and the point \((1, 2)\).

\(f = x^{2} y,\; \mathbf{v} = \langle 3, 4 \rangle,\; P = (1, 2)\)

Step 2 - Find the gradient by taking the partial derivatives.

In this problem: The gradient is \(\nabla f = \langle 2 x y, x^{2} \rangle\).

\(\nabla f = \left\langle 2 x y,\; x^{2} \right\rangle\)

Step 3 - Turn the direction into a unit vector.

In this problem: The magnitude is \(\lVert \mathbf{v} \rVert = 5\), so the unit vector is \(\mathbf{u} = \langle \frac{3}{5}, \frac{4}{5} \rangle\).

\(\mathbf{u} = \left\langle \frac{3}{5},\; \frac{4}{5} \right\rangle\)

Step 4 - Dot the gradient with the unit vector.

In this problem: Dotting the gradient with the unit vector gives \(D_{\mathbf{u}} f = \frac{2 x \left(2 x + 3 y\right)}{5}\).

\(D_{\mathbf{u}} f = \frac{2 x \left(2 x + 3 y\right)}{5}\)

Step 5 - Evaluate at the given point.

In this problem: Evaluating at \((1, 2)\) gives \(\frac{16}{5}\).

\(D_{\mathbf{u}} f(1, 2) = \frac{16}{5}\)

Final answer:

\(D_{\mathbf{u}} f = \frac{16}{5}\)

Example 2:

\(f = x^{2} + y^{2},\; \mathbf{v} = \langle 1, 1 \rangle,\; P = (1, 2)\)

Step 1 - Set up the function, the direction vector, and the point.

In this problem: We are given \(f = x^{2} + y^{2}\), direction \(\mathbf{v} = \langle 1, 1 \rangle\), and the point \((1, 2)\).

\(f = x^{2} + y^{2},\; \mathbf{v} = \langle 1, 1 \rangle,\; P = (1, 2)\)

Step 2 - Find the gradient by taking the partial derivatives.

In this problem: The gradient is \(\nabla f = \langle 2 x, 2 y \rangle\).

\(\nabla f = \left\langle 2 x,\; 2 y \right\rangle\)

Step 3 - Turn the direction into a unit vector.

In this problem: The magnitude is \(\lVert \mathbf{v} \rVert = \sqrt{2}\), so the unit vector is \(\mathbf{u} = \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \rangle\).

\(\mathbf{u} = \left\langle \frac{\sqrt{2}}{2},\; \frac{\sqrt{2}}{2} \right\rangle\)

Step 4 - Dot the gradient with the unit vector.

In this problem: Dotting the gradient with the unit vector gives \(D_{\mathbf{u}} f = \sqrt{2} \left(x + y\right)\).

\(D_{\mathbf{u}} f = \sqrt{2} \left(x + y\right)\)

Step 5 - Evaluate at the given point.

In this problem: Evaluating at \((1, 2)\) gives \(3 \sqrt{2} \approx 4.24\).

\(D_{\mathbf{u}} f(1, 2) = 3 \sqrt{2} \approx 4.24\)

Final answer:

\(D_{\mathbf{u}} f = 3 \sqrt{2} \approx 4.24\)