Directional Derivative Calculator

Published on: July 19, 2026

This Directional Derivative Calculator helps you find the directional derivative of a multivariable function at a given point. The calculator first finds the gradient vector, then converts the given direction into a unit vector, and finally takes their dot product to get the result. This shows the rate of change of the function in the chosen direction. It is a simple way to check answers, understand the method clearly, and practise multivariable calculus step by step.

Step-by-step method

  1. Identify the function, the point, and the direction vector.
  2. Compute each partial derivative with substeps.
  3. Combine them into the gradient vector ∇f.
  4. Evaluate ∇f at the point.
  5. Convert the direction vector into a unit vector.
  6. Use Dûf = ∇f(a) · û.

Formulas:

Gradient vector formula

2D: ∇f = ⟨ fx, fy
3D: ∇f = ⟨ fx, fy, fz

Unit direction vector formula

û =
v
|v|
2D: |v| = a2 + b2
3D: |v| = a2 + b2 + c2

Directional derivative formula

Dûf( a ) = ∇f( a ) · û

Example 1: f(x,y)=yx^2+sin(xy); (1,2); <3,4>

Step 1A - Identify the function.

In this problem: Read the given function.

f(x, y) = yx2 + sin(xy)

Step 1B - Identify the point.

In this problem: Read the point where the directional derivative is required.

Point = ( 1, 2 )

Step 1C - Identify the direction vector.

In this problem: Read the given direction vector.

v = ⟨ 3, 4 ⟩

Step 2A - Differentiate term 1 with respect to x.

In this problem: Rule shown by the math.

∂x
(yx2) = y·
∂x
(x2)

Step 2B - Differentiate term 1 with respect to x.

In this problem: Rule shown by the math.

∂x
(x2) = 2x

Step 2C - Differentiate term 1 with respect to x.

In this problem: Rule shown by the math.

∂x
(yx2) = 2xy

Step 2D - Differentiate term 2 with respect to x.

In this problem: Rule shown by the math.

∂x
(sin(xy)) = cos(xy)·
∂x
(xy)

Step 2E - Differentiate term 2 with respect to x.

In this problem: Rule shown by the math.

∂x
(xy) = y

Step 2F - Differentiate term 2 with respect to x.

In this problem: Rule shown by the math.

cos(xy)·y = y·cos(xy) = y·cos(xy)

Step 2G - Combine the results.

In this problem: Add the derivatives from the previous substeps.

fx = 2xy + y·cos(xy) = y(2x + cos(xy))

Step 3A - Differentiate term 1 with respect to y.

In this problem: Rule shown by the math.

∂y
(yx2) = x2·
∂y
(y)

Step 3B - Differentiate term 1 with respect to y.

In this problem: Rule shown by the math.

∂y
(y) = 1

Step 3C - Differentiate term 1 with respect to y.

In this problem: Rule shown by the math.

∂y
(yx2) = x2

Step 3D - Differentiate term 2 with respect to y.

In this problem: Rule shown by the math.

∂y
(sin(xy)) = cos(xy)·
∂y
(xy)

Step 3E - Differentiate term 2 with respect to y.

In this problem: Rule shown by the math.

∂y
(xy) = x

Step 3F - Differentiate term 2 with respect to y.

In this problem: Rule shown by the math.

cos(xy)·x = x·cos(xy) = x·cos(xy)

Step 3G - Combine the results.

In this problem: Add the derivatives from the previous substeps.

fy = x2 + x·cos(xy) = x(x + cos(xy))

Step 4 - Combine into the gradient vector.

In this problem: Put the partial derivatives into ⟨ … ⟩.

∇f = ⟨ y(2x + cos(xy)), x(x + cos(xy)) ⟩

Step 5A - Substitute the point into fx.

In this problem: Substitute the point values.

fx( 1, 2 ) = 2(2 + cos( 2 ))

Step 5B - Simplify fx at the point.

In this problem: Simplify the substituted expression.

fx( 1, 2 ) = 2(2 + cos( 2 )) ≈ 3.17

Step 5C - Substitute the point into fy.

In this problem: Substitute the point values.

fy( 1, 2 ) = 1 + cos( 2 )

Step 5D - Simplify fy at the point.

In this problem: Simplify the substituted expression.

fy( 1, 2 ) = 1 + cos( 2 ) ≈ 0.58

Step 6 - Gradient at the point.

In this problem: Put the evaluated components into ⟨ … ⟩.

∇f( 1, 2 ) = ⟨ 3.17, 0.58 ⟩

Step 7A - Find the magnitude of the direction vector.

In this problem: Use the vector magnitude formula.

|v| = 32 + 42

Step 7B - Simplify the magnitude.

In this problem: Simplify the square root.

|v| = 5

Step 8 - Convert the direction vector into a unit vector.

In this problem: Divide the direction vector by its magnitude.

û =
⟨ 3, 4 ⟩
5
=
3
5
,
4
5

Step 9A - Use the directional derivative formula.

In this problem: Substitute the gradient at the point and the unit vector.

Dûf( 1, 2 ) = ∇f( 1, 2 ) · û = ⟨ 3.17, 0.58 ⟩ ·
3
5
,
4
5

Step 9B - Expand the dot product.

In this problem: Multiply matching components and add.

Dûf( 1, 2 ) = 2(2 + cos( 2 ))·
3
5
+ 1 + cos( 2 )·
4
5

Step 9C - Simplify the result.

In this problem: Write the final directional derivative.

Dûf( 1, 2 ) =
16 + 10·cos( 2 )
5
≈ 2.37

Final answer: Directional derivative = 2*cos(2) + 16/5 ≈ 2.37

Example 2: f(x,y)=x^2+y^2; (1,1); <1,1>

Step 1A - Identify the function.

In this problem: Read the given function.

f(x, y) = x2 + y2

Step 1B - Identify the point.

In this problem: Read the point where the directional derivative is required.

Point = ( 1, 1 )

Step 1C - Identify the direction vector.

In this problem: Read the given direction vector.

v = ⟨ 1, 1 ⟩

Step 2A - Differentiate term 1 with respect to x.

In this problem: Rule shown by the math.

∂x
(x2) = 2x

Step 2B - Differentiate term 2 with respect to x.

In this problem: Rule shown by the math.

∂x
(y2) = 0

Step 2C - Combine the results.

In this problem: Add the derivatives from the previous substeps.

fx = 2x + 0 = 2x

Step 3A - Differentiate term 1 with respect to y.

In this problem: Rule shown by the math.

∂y
(x2) = 0

Step 3B - Differentiate term 2 with respect to y.

In this problem: Rule shown by the math.

∂y
(y2) = 2y

Step 3C - Combine the results.

In this problem: Add the derivatives from the previous substeps.

fy = 0 + 2y = 2y

Step 4 - Combine into the gradient vector.

In this problem: Put the partial derivatives into ⟨ … ⟩.

∇f = ⟨ 2x, 2y ⟩

Step 5A - Substitute the point into fx.

In this problem: Substitute the point values.

fx( 1, 1 ) = 2

Step 5B - Simplify fx at the point.

In this problem: Simplify the substituted expression.

fx( 1, 1 ) = 2

Step 5C - Substitute the point into fy.

In this problem: Substitute the point values.

fy( 1, 1 ) = 2

Step 5D - Simplify fy at the point.

In this problem: Simplify the substituted expression.

fy( 1, 1 ) = 2

Step 6 - Gradient at the point.

In this problem: Put the evaluated components into ⟨ … ⟩.

∇f( 1, 1 ) = ⟨ 2, 2 ⟩

Step 7A - Find the magnitude of the direction vector.

In this problem: Use the vector magnitude formula.

|v| = 12 + 12

Step 7B - Simplify the magnitude.

In this problem: Simplify the square root.

|v| = 2 ≈ 1.41

Step 8 - Convert the direction vector into a unit vector.

In this problem: Divide the direction vector by its magnitude.

û =
⟨ 1, 1 ⟩
2
=
2
2
,
2
2

Step 9A - Use the directional derivative formula.

In this problem: Substitute the gradient at the point and the unit vector.

Dûf( 1, 1 ) = ∇f( 1, 1 ) · û = ⟨ 2, 2 ⟩ ·
2
2
,
2
2

Step 9B - Expand the dot product.

In this problem: Multiply matching components and add.

Dûf( 1, 1 ) = 2·
2
2
+ 2·
2
2

Step 9C - Simplify the result.

In this problem: Write the final directional derivative.

Dûf( 1, 1 ) = 22 ≈ 2.83

Final answer: Directional derivative = 2*sqrt(2) ≈ 2.83