[Maths Class Notes] on Directional Derivative Pdf for Exam

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)

Directional Derivative Definition

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)

How to Find The Directional Derivative?

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)

Directional derivative properties

Some basic directional derivative properties are as follows:

1. The rule for a constant factor

▽v (pf) = p▽vf

2. Rules for the Sum

▽v (f + h) =▽vf + ▽vh

3. Rules for the product.

The rule for products is also known as Leibniz rule.

▽v (fh) = h▽vf + f▽vh

4. 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)

Directional Derivative Formula

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

Solved Examples

1. 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}]

2. 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

Quiz Time

1. 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. 1 2(-i + j)

2. 1 2(i – j)

3. 1 2( i + j)

4. 1 5( 2i + j)

5. -1 5(i – j)

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

1. 1 4i + 1 2j

2. -96i – 56j

3. -152

4. -30.4

5. -32i + 14j