Directional Derivative Calculator

What is a Directional Derivative Calculator?

A Directional Derivative Calculator evaluates the instantaneous rate of change of a scalar field \(f(x,y[,\!z])\) at a point in a specified direction. Unlike partial derivatives, which measure change along coordinate axes, the directional derivative measures change along an arbitrary direction vector. If \(\widehat{\mathbf{u}}\) is a unit direction, the directional derivative at point \(\mathbf{p}\) is \[ D_{\widehat{\mathbf{u}}}f(\mathbf{p})=\nabla f(\mathbf{p})\cdot \widehat{\mathbf{u}}, \] where \(\nabla f\) is the gradient. Geometrically, \(D_{\widehat{\mathbf{u}}}f\) is the slope of the tangent plane to the graph of \(f\) in the direction \(\widehat{\mathbf{u}}\). Its magnitude is maximized in the gradient direction, with maximum value \(\|\nabla f(\mathbf{p})\|\), and minimized in the opposite direction, \(-\nabla f\). The calculator automates normalization, gradient computation, substitution, and dot products, and can also provide a symmetric finite-difference approximation when symbolic differentiation is not available.

About the Directional Derivative Calculator

Given a nonunit direction \(\mathbf{v}\), the calculator first forms the unit vector \[ \widehat{\mathbf{u}}=\frac{\mathbf{v}}{\|\mathbf{v}\|},\qquad \|\mathbf{v}\|=\sqrt{v_1^2+v_2^2[+v_3^2]}. \] It then computes the gradient \[ \nabla f=\left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}[,\,\frac{\partial f}{\partial z}] \right\rangle, \] evaluates \(\nabla f(\mathbf{p})\), and returns \(D_{\widehat{\mathbf{u}}}f(\mathbf{p})=\nabla f(\mathbf{p})\cdot \widehat{\mathbf{u}}\). By the angle formula, \[ D_{\widehat{\mathbf{u}}}f(\mathbf{p})=\|\nabla f(\mathbf{p})\|\cos\theta, \] where \(\theta\) is the angle between \(\nabla f(\mathbf{p})\) and \(\widehat{\mathbf{u}}\). For a parameterized curve \(\boldsymbol{\gamma}(t)\) through \(\mathbf{p}\) with velocity \(\boldsymbol{\gamma}'(t_0)\), \[ \frac{d}{dt}f(\boldsymbol{\gamma}(t))\Big|_{t_0}=\nabla f(\mathbf{p})\cdot \boldsymbol{\gamma}'(t_0), \] and the directional derivative along the curve’s unit tangent is this quantity divided by \(\|\boldsymbol{\gamma}'(t_0)\|\). A robust numeric estimate is the central difference \[ D_{\widehat{\mathbf{u}}}f(\mathbf{p})\approx \frac{f(\mathbf{p}+h\widehat{\mathbf{u}})-f(\mathbf{p}-h\widehat{\mathbf{u}})}{2h},\quad h\ \text{small}. \]

How to Use this Directional Derivative Calculator

1. Enter the scalar field \(f(x,y[,\!z])\) and the evaluation point \(\mathbf{p}=(x_0,y_0[,\!z_0])\).
2. Enter the direction vector \(\mathbf{v}\) (any nonzero vector). The tool normalizes it to \(\widehat{\mathbf{u}}\).
3. Compute the gradient, substitute \(\mathbf{p}\), and form the dot product \(\nabla f(\mathbf{p})\cdot \widehat{\mathbf{u}}\).
4. (Optional) Use the finite-difference preview to cross-check the symbolic result.
5. Review the steps and copy the LaTeX expressions for your notes or report.

Core Formulas (LaTeX)

Unit direction & gradient: \[ \widehat{\mathbf{u}}=\frac{\mathbf{v}}{\|\mathbf{v}\|},\qquad \nabla f=\left\langle f_x,f_y[,f_z]\right\rangle. \]

Directional derivative: \[ D_{\widehat{\mathbf{u}}}f(\mathbf{p})=\nabla f(\mathbf{p})\cdot \widehat{\mathbf{u}}. \]

Angle relation (steepest ascent): \[ D_{\widehat{\mathbf{u}}}f(\mathbf{p})=\|\nabla f(\mathbf{p})\|\cos\theta,\quad \max= \|\nabla f(\mathbf{p})\|\ \text{when }\widehat{\mathbf{u}}\parallel \nabla f. \]

Central difference estimate: \[ D_{\widehat{\mathbf{u}}}f(\mathbf{p})\approx\frac{f(\mathbf{p}+h\widehat{\mathbf{u}})-f(\mathbf{p}-h\widehat{\mathbf{u}})}{2h}. \]

Examples (Illustrative)

Example 1 — 2D polynomial

\(f(x,y)=x^2y+y^3\), \(\mathbf{p}=(1,2)\), \(\mathbf{v}=(3,4)\Rightarrow \widehat{\mathbf{u}}=(\tfrac{3}{5},\tfrac{4}{5})\). \(\nabla f=(2xy,\ x^2+3y^2)\Rightarrow (4,13)\) at \((1,2)\). \(D_{\widehat{\mathbf{u}}}f=(4,13)\cdot(\tfrac{3}{5},\tfrac{4}{5})=\tfrac{12}{5}+\tfrac{52}{5}= \tfrac{64}{5}=12.8.\)

Example 2 — 3D exponential-quadratic

\(f(x,y,z)=e^{xz}+y^2\), \(\mathbf{p}=(0,1,2)\), \(\mathbf{v}=(1,-1,2)\). \(\nabla f=(z e^{xz},\ 2y,\ x e^{xz})\Rightarrow (2,2,0)\). Since \((2,2,0)\cdot(1,-1,2)=0\), the directional derivative is \(0\) (direction orthogonal to the gradient).

Example 3 — Along a curve’s tangent

\(f(x,y)=x^2+y^2\), curve \(\boldsymbol{\gamma}(t)=(t,t^2)\) at \(t=1\Rightarrow \mathbf{p}=(1,1)\), \(\boldsymbol{\gamma}'(1)=(1,2)\), \(\widehat{\mathbf{u}}=\tfrac{1}{\sqrt{5}}(1,2)\). \(\nabla f=(2x,2y)\Rightarrow (2,2)\). \(D_{\widehat{\mathbf{u}}}f=\frac{(2,2)\cdot(1,2)}{\sqrt{5}}=\frac{6}{\sqrt{5}}\approx2.683.\)

FAQs

What’s the difference between a directional derivative and the gradient?

The gradient is a vector; a directional derivative is its dot product with a unit direction, giving a scalar rate.

Does the direction have to be a unit vector?

Yes. Always normalize: \(D_{\widehat{\mathbf{u}}}f=\nabla f\cdot\frac{\mathbf{v}}{\|\mathbf{v}\|}\).

How do directional and partial derivatives relate?

Partials are directions along axes; e.g., \(f_x=D_{\mathbf{e}_x}f\), \(f_y=D_{\mathbf{e}_y}f\).

What if \(f\) isn’t differentiable at the point?

The gradient may not exist; some one-sided directional limits may exist but can be inconsistent.

How should I choose the step \(h\) for numeric estimates?

Use a small \(h\); too small can amplify roundoff. Try \(10^{-4}\)–\(10^{-6}\) and check stability.

Why is my directional derivative zero?

Your direction is orthogonal to the gradient; the function doesn’t change to first order in that direction.