The directional derivative is the rate at which any function changes at any specific point in a fixed direction. It is considered as a vector form of any derivative. It specifies the immediate rate of variation of the function. It particularised the vision of partial derivatives. It can be represented as :

▽uf = ▽f .( u/|u|)

= limh→0[f(k + h û) –f(k)]/h

In this article, we will discuss the concept of directional derivative in detail. We will study what is directional derivative, directional derivative definition, how to find the directional derivative, directional derivative formula, directional derivative properties etc.

\[D_{u}\] f(x,y)

For a scalar function f(k) =f (k₁ , k₂,....kn), the directional derivative is defined as a function in the following manner,

▽uf = limh→0[f(k + hv) –f(k)]/h

Where v is considered as a vector along which the directional derivative f(k) is defined. Sometimes v is confined to a unit vector, or else, the definition also holds.

Vector v is derived by,

V = (v₁ , v₂,....vn)

The first step to find the directional derivative is to mention the direction. One method to mention the direction is with a vector u ( u₁ , u₂) that points in the direction in which we wish to find the slope. We will consider u as a unit vector. Using the directional derivative definition, we can find the directional derivative f at k in the direction of a unit vector u as

Duf (k). We can define it with a limit definition just as a standard derivative or partial derivative.

Du f (k) = limh→0[f(k +hu) –f(k)]/h

The concept of directional derivatives is quite easy to understand. Du f (k) is the slope of f(x,y) when standing at the point k and facing the direction by a unit vector (u). x and y are represented in meters then Du f (k) will be changed in height per meter as you move in the direction given by u when you are standing at the point k.

Note: Du f (k) is a matrix not a number. Directional derivative is similar as a partial derivative if u points in the positive x or positive y direction. For example if u= (1,0) then

Du f (k) = \[\partial\] f/\[\partial\]x (k). Similarly if unit vector (u) = (0,1) then,

Du f (k) = \[\partial\]f/\[\partial\]x (k)

Some basic directional derivative properties are as follows:

The rule for a constant factor

▽v (pf) = p▽vf

Rules for the Sum

▽v (f + h) =▽vf + ▽vh

Rules for the product.

The rule for products is also known as Leibniz rule.

▽v (fh) = h▽vf + f▽vh

Chain rule

The chain rule is used when function f is differentiable at ‘a’ and g is differentiable at f(a). In such a case,

▽v ( f o h) (a) = f’(h(a)) ▽vh(a)

The directional derivative formula is represented as n.▽f. Here, n is considered as a unit vector. The directional derivative is stated as the rate of change along with the path of the unit vector which is u =(p,q). The directional derivative is represented by Du F(p,q) which can be written as follows:

Du f (p,q) = limh→0[f(x + ph, y +qh) –f(p,q)]/h

For the function f(m,n) = m²n., find the directional derivative of f at the point (3,2) in the direction of (2,1).

Solution: The unit vector in the direction of (2,1)

u = (2,1)/\[\sqrt{5}\]= (2/\[\sqrt{5}\], 1/\[\sqrt{5}\])

Since.,we are at the point (3,2), ( equation1) is still valid. Now we will use another value of the unit vector to get.

DU f (3, 2) = 12u1 + 9u2

= 24/\[\sqrt{5}\] + 9/\[\sqrt{5}\] = 33/\[\sqrt{5}\]

Find the directional derivative of the function f(p,q) = pqr in the direction 3i-4k. It has the point as (1,-1,1).

Solution:

Given function f(p,q) = pqr

Vector field is 3i - 4k. It has the magnitude of \[\sqrt{(3^{2}) +(-4^{2})}\] = \[\sqrt{25}\]= \[\sqrt{5}\]

The unit vector n in the direction 3i - 4k is n = 1/5(3i- 4k)

Now,we have to calculate the gradient ▽ f for calculating the directional derivative.

Hence,▽ f = qri +pri + pqk

Now, the directional derivative is

n▽ f = ⅕(3i-4k).( qri +pri + pqk)

= ⅕[ 3 × qr + 0- 4 * pq)

The directional derivative at the point (1,-1,1) is

n.▽ f = 1/5[ 3 × (-1) × (1) - 4 ×1 × (-1)

n.▽ f = 1/5

Find the direction in which which the directional derivative is greater for the function

f(m,n) = 3m² 2n² - m⁴ -n⁴ at the point (1,2).

1 2(-i + j)

1 2(i - j)

1 2( i + j)

1 5( 2i + j)

-1 5(i - j)

2. The directional derivative f(m,n) = m²n³ - 2m4n at the point (1,2) in the direction 3i-4j.

1 4i + 1 2j

-96i - 56j

-152

-30.4

-32i + 14j

FAQ (Frequently Asked Questions)

1. Explain Partial Derivative.

Suppose we have two variables f(m,n), which relies on two variables i.e. m and n, where m and n are independent of each other. Then we can say that function f is partially dependent on m and n,. Now, if we calculate the derivative of f, then that derivative is called a partial derivative. If we differentiate the function f in terms of m, then we will consider n as constant and if we differentiate f in terms of y, then variable m will be taken as constant.

The partial derivative of function f in terms of m is differently represented by f’m, f m , ∂f or ∂f/m. Here, the partial derivative is represented by the symbol ∂.

2. Explain Unit vector.

A unit vector is defined as a vector that has a magnitude of 1 unit. It is denoted by a lowercase letter with a cap (“ˆ”) a vector in space is represented by unit vectors. Any vector can change into a unit vector by dividing it through the magnitude of the vector are as follows:

Unit Vector = Vector/Magnitude of the vector

A unit vector having a similar direction as the vector ˆm is derived as:

᷍m= m̅ |m̅|

Here,

᷍m = Denotes a unit vector

m̅ = Denotes the vector

|m̅| = Denotes the magnitude of the vector

Note: Any two unit vectors such as ˆm and ˆn are not considered as unit vectors because they have similar magnitude. As the directions in which vectors are derived may be different because these unit vectors are different from each other. Hence, to define a vector both magnitude and directions should be given.