<< Chapter < Page Chapter >> Page >

Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.

Directional derivative of a function of two variables

Let z = f ( x , y ) be a function of two variables x and y , and assume that f x and f y exist. Then the directional derivative of f in the direction of u = cos θ i + sin θ j is given by

D u f ( x , y ) = f x ( x , y ) cos θ + f y ( x , y ) sin θ .

Proof

[link] states that the directional derivative of f in the direction of u = cos θ i + sin θ j is given by

D u f ( a , b ) = lim t 0 f ( a + t cos θ , b + t sin θ ) f ( a , b ) t .

Let x = a + t cos θ and y = b + t sin θ , and define g ( t ) = f ( x , y ) . Since f x and f y both exist, we can use the chain rule for functions of two variables to calculate g ( t ) :

g ( t ) = f x d x d t + f y d y d t = f x ( x , y ) cos θ + f y ( x , y ) sin θ .

If t = 0 , then x = x 0 and y = y 0 , so

g ( 0 ) = f x ( x 0 , y 0 ) cos θ + f y ( x 0 , y 0 ) sin θ .

By the definition of g ( t ) , it is also true that

g ( 0 ) = lim t 0 g ( t ) g ( 0 ) t = lim t 0 f ( x 0 + t cos θ , y 0 + t sin θ ) f ( x 0 , y 0 ) t .

Therefore, D u f ( x 0 , y 0 ) = f x ( x , y ) cos θ + f y ( x , y ) sin θ .

Finding a directional derivative: alternative method

Let θ = arccos ( 3 / 5 ) . Find the directional derivative D u f ( x , y ) of f ( x , y ) = x 2 x y + 3 y 2 in the direction of u = ( cos θ ) i + ( sin θ ) j . What is D u f ( −1 , 2 ) ?

First, we must calculate the partial derivatives of f :

f x = 2 x y f y = x + 6 y ,

Then we use [link] with θ = arccos ( 3 / 5 ) :

D u f ( x , y ) = f x ( x , y ) cos θ + f y ( x , y ) sin θ = ( 2 x y ) 3 5 + ( x + 6 y ) 4 5 = 6 x 5 3 y 5 4 x 5 + 24 y 5 = 2 x + 21 y 5 .

To calculate D u f ( −1 , 2 ) , let x = −1 and y = 2 :

D u f ( −1 , 2 ) = 2 ( −1 ) + 21 ( 2 ) 5 = −2 + 42 5 = 8 .

This is the same answer obtained in [link] .

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the directional derivative D u f ( x , y ) of f ( x , y ) = 3 x 2 y 4 x y 3 + 3 y 2 4 x in the direction of u = ( cos π 3 ) i + ( sin π 3 ) j using [link] . What is D u f ( 3 , 4 ) ?

D u f ( x , y ) = ( 6 x y 4 y 3 4 ) ( 1 ) 2 + ( 3 x 2 12 x y 2 + 6 y ) 3 2 D u f ( 3 , 4 ) = 72 256 4 2 + ( 27 576 + 24 ) 3 2 = −94 525 3 2

Got questions? Get instant answers now!

If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in [link] in the direction of the vector −5 , 12 , we would first divide by its magnitude to get u . This gives us u = ( 5 / 13 ) , 12 / 13 . Then

D u f ( x , y ) = f ( x , y ) · u = 5 13 ( 2 x y ) + 12 13 ( x + 6 y ) = 22 13 x + 17 13 y .

Gradient

The right-hand side of [link] is equal to f x ( x , y ) cos θ + f y ( x , y ) sin θ , which can be written as the dot product of two vectors. Define the first vector as f ( x , y ) = f x ( x , y ) i + f y ( x , y ) j and the second vector as u = ( cos θ ) i + ( sin θ ) j . Then the right-hand side of the equation can be written as the dot product of these two vectors:

D u f ( x , y ) = f ( x , y ) · u .

The first vector in [link] has a special name: the gradient of the function f . The symbol is called nabla and the vector f is read “del f .”

Definition

Let z = f ( x , y ) be a function of x and y such that f x and f y exist. The vector f ( x , y ) is called the gradient    of f and is defined as

f ( x , y ) = f x ( x , y ) i + f y ( x , y ) j .

The vector f ( x , y ) is also written as “grad f .”

Finding gradients

Find the gradient f ( x , y ) of each of the following functions:

  1. f ( x , y ) = x 2 x y + 3 y 2
  2. f ( x , y ) = sin 3 x cos 3 y

For both parts a. and b., we first calculate the partial derivatives f x and f y , then use [link] .


  1. f x ( x , y ) = 2 x y and f y ( x , y ) = x + 6 y , so f ( x , y ) = f x ( x , y ) i + f y ( x , y ) j = ( 2 x y ) i + ( x + 6 y ) j .

  2. f x ( x , y ) = 3 cos 3 x cos 3 y and f y ( x , y ) = −3 sin 3 x sin 3 y , so f ( x , y ) = f x ( x , y ) i + f y ( x , y ) j = ( 3 cos 3 x cos 3 y ) i ( 3 sin 3 x sin 3 y ) j .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the gradient f ( x , y ) of f ( x , y ) = ( x 2 3 y 2 ) / ( 2 x + y ) .

f ( x , y ) = 2 x 2 + 2 x y + 6 y 2 ( 2 x + y ) 2 i x 2 + 12 x y + 3 y 2 ( 2 x + y ) 2 j

Got questions? Get instant answers now!
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask