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Directional derivative is the rate at which any function changes at any specific point in a fixed direction. The concept of the directional derivative is simple; Duf(a) is the slope of f (x, y) when you stand in position a and look at the direction given to u. If x and y were given in meters, then Duf (a) would be the change in height per meter as you go from place to place in u when you are in place a. It is considered a vector method of any derivative. Specifies the fastest level of job variability. It defines the vision of partial derivatives.
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Key Takeaways: Directional derivative, Derivative, Vector, Unit vector, Chain rule, Product rule
Also read: Isosceles Triangle Theorems
Directional Derivative Definition
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In the scale function f (k) = f (k1, k2, .... kn), the direct output is defined as the function as follows,
∇uf = limh→0f(k+hv)–f(k)f(k+hv)–f(k)/h
Where v is considered a vector where the output of the other f (k) is defined. Sometimes the v is locked in the unit vector, or, the definition also holds.
Vector v taken by,
V = (v1 , v2,....vn)
(OR)
The rate of change of f(x,y)f(x,y) in the direction of the unit vector →u=⟨a,b⟩ u→=⟨a,b⟩ is called the directional derivative and is denoted by D→u f(x,y) Du→ f(x,y). The definition of the directional derivative is,
D→u f(x,y) = limh→0 [f (x + ah , y + bh) − f(x,y)] / h
Methods to Find Directional Derivatives
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The first step in getting out of the other direction is to say the direction. Another way to say direction is with the vector u (u1, u2) which indicates where we wish to find the slope. We will take you as a unit vector. Using the definition of a direct derivative, we can find the direct output of f in k for each vector of unit u as
Duf (k). We can define a definition of a boundary as a standard exit or partial output.
Du f (k) = limh→0 f (k + hu) – f(k) f(k + hu) – f(k)/h
The concept of direct exit is easy to understand. Du f(k) is the slope of f (x, y) when you stand in position k and look at the direction of the vector unit (u). x and y are represented by meters and Du f(k) will be changed in length by each meter as you move to the side given by u when standing in position k.
Du f(k) = ∂ f/∂x (k).
Similarly, if it is a vector unit, (u) = (0,1) then,
Du f(k) = ∂ f/∂x (k).
Out of the other direction del _ (u) f (x0, y0, z0) is the rate at which the function f (x, y, z) changes in a certain area (x0, y0, z0) on side u. It is a type of vector output of normal, and can be defined as
del (u) f = del f·(u)/(|u|)
= lim (f(x+hu^^)-f(x))/h,
(h 0)
where is called "nabla" or "del" and u^^ denotes a unit vector.
Directional Derivative is also written as:
d/(ds) =s^^·del
where s means unit vector in any direction and partial f/partial x=fx means output from the other part.
Also read: First Order Differential Equation
Directional Derivative Properties
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- The rule for a constant factor
∇v (pf) = p∇vf
- Rules for the Sum
∇v (f + h) =∇vf +∇vh
- Rules for the product.
This rule is also known as Leibniz Rule
∇v (fh) = h∇vf + f∇vh
- Chain rule
∇v ( f o h) (a) = f’(h(a)) ∇vh(a)
Directional Derivative Formula
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The directional derivative formula is represented as n. ∇ f. Here, n is considered the unit vector. The output is defined as the change rate and the unit vector method u = (p, q). The alternative direction is represented by Du F (p, q) which can be written as follows:
Du f (p,q) = limh→0f(x + p h, y + q h)–f( p , q ) f (x + p h , y + q h )–f( p , q ) / h
Also Read: Discrete Maths
Points To Remember
- The directional derivative of f at the point (x , y) in the direction of the unit vector u = ⟨u1 , u2⟩ is: Du f (x , y) = limh→0 {f( x + u1h, y + u2h) – f ( x , y )}/h
- The directional derivative in the direction u may be computed by: Du f(x0 , y0) = ∇ f(x0 , y0)⋅u.
- Directional derivative helps in understanding the multivariable function changes when it moves within some vectors in its input space.
- The rule for a constant factor - ∇v (pf) = p∇vf
- Rules for the sum – ∇v (f + h) =∇vf +∇vh
- Rules for the product - ∇v (fh) = h∇vf + f∇vh
- Chain rule - ∇v ( f o h) (a) = f’(h(a)) ∇vh(a)
Also Read: Absolute values
Sample Questions
Ques. With the function f (m, n) = m2n., Find the output of the other direction f in the area (3,2) in the direction of (2,1). (2 marks)
Ans: Unit vector toward (2,1)
u = (2,1) / √5
= (2/√5, 1/√5)
As., We are at point (3,2), (number1) still active. We will now use another unit vector value to obtain it.
DU f (3, 2) = 12u1 + 9u2
= 24/√5 + 9/√5
= 33/√5
Ques. Find out the direction of f in P position on the indicated side. f(x, y)=x2lny, P(5, 1), a=−i+4j (5 marks)
Ans: ∇f=(∂f∂x)i+(∂f∂y)j+(∂f∂z)k
The derivative of a function at a point P(x0,y00,z0) in the direction of a given vector, u is given as:
Dv=∇f(x0,y0,z0)⋅v|/|v|
f(x, y) = x2lny
∇f(x,y) = ∂/∂xf(x, y)i + ∂/∂yf(x, y)j
∇f(x,y) = ∂/∂x(x2lny)i + ∂∂y(x2lny)j
∇f(x,y) = 2xlnyi + x2⋅1yj
∇f(5,1) = 2(5)ln(1)i + (5)2⋅1(1)j
∇f(5,1) = 0i + (25)⋅1j
∇f(5,1)| = √(0)2 +(25)2
|∇f(5,1)| = 25
The output of each of the given vector functions is equal to the product output scale of the function gradient and the unit vector.
So, now let's find out the direction from the given vectors:
∇f(5,1)=25j
a=−i+4j
Du = ∇f(x0,y0)⋅a/|→a|
=(25j)⋅(−i+4j)/|−i+4j|
=0+(25)(4)/√(−1)2+(4)2
=100/√17
So, Directional Derivative of the function id Du = 100/√7
Ques. Let θ=arccos(3/5).Find the directional derivative Du → f (x,y) Du → f (x,y) of f(x,y) = x2 − xy + 3y2 f (x,y) = x2 − xy + 3y2 in the direction of u→=(cosθ)i^ + (sinθ)ju. Then determine Du→f(−1,2)Du→f(−1,2). (4 marks)
Ans: Calculate Partial Derivative of f:
fx(x,y) =2x−y
fy(x,y) =−x+6y,
Du→f(x,y)=fx(x,y)cosθ+fy(x,y)sinθ
=(2x−y)3/5+(−x+6y)4/5
=6x/5−3y/5−4x/5+24y/5
=(2x+21y)/5
Now, to calculate Duf(-1,2), let x=-1 and y=2:
Duf(-1,2) = [2(-1)+21(2)]/5
= (-2+42)/5
= 8
Ques. Find the Directional Derivative of f(x,y,z) = xyz at point (-1,1,3) in the direction of vector i-2j+2k is? (1 marks)
Ans: 7/3
Ques. Calculate Du? f(x,y,z), where f(x,y,z)=x2y+xz10+xy2z5 in the direction of v?(1,2,3). (2 marks)
Ans: ||\(\overrightarrow {v}\)||=√(12+22+32) = √(1+4+9) = √14≠1
\(\overrightarrow {u}\) =1/√14\(\overrightarrow {v}\) = (1/√14, 2/√14, 3/√14)
Now, do a partial derivative and multiply each vector corresponding to it.
D\(\overrightarrow {u}\) f(x,y,z)= fx (x,y,z) a+ fy (x,y,z) b+ fz (x,y,z)c
D\(\overrightarrow {u}\) f(x,y,z)=1/√14 (2xy+z10+y2z5)+2/√14(x2+0+2xyz5)+3/√14(0+10xz9+5xy2z4)
D\(\overrightarrow {u}\) f(x,y,z)=1/√14√(2xy+z10+y2z5)+2/√14(x2+2xyz5)+3/√14(10xz9+5xy2z4)
Du? f(x,y,z)=1/√14((2xy+z10+y2z5)+2(x2+2xyz5)+3(10xz9+5xy2z4))
Ques. If f ( x , y ) = 3x2−1y2, find the value of the directional derivative at the point (−1,−4) in the direction given by the angle θ = 2π/6 (1 marks)
Ans: 3.92820323027551
Ques. Find the directional derivative of the function f( x,y, z) = 3 x y + z2 at the point ( 1 , −2 , 2) in the direction from that point toward the origin. (3 marks)
Ans: Vector from that point to the origin:
v = (−1, 2, −2)
Unit vector on location:
u = {1 /|v|} * v = ( −1/3 , 2/3 , −2/3)
Directional derivative in direction u:
Du f (1 , − 2 , 2) = ∇f ( 1, −2, 2) . u
∇f (x , y, z) = (3 y, 3 x, 2z) ∇f( 1 , −2, 2 ) = (−6 , 3, 4)
Du f (1 , −2, 2) = ∇f(1 , −2 , 2 ) . u = (−6 , 3, 4) ·( −1/3 , 2/3 , −2/3) = 4/3
Ques. Let f( x, y, z ) = x2 − y2 + xyz and v = (3, 4, 12). Find the directional derivative of f in the direction of the vector v at the point (1 , 2 , −1). (3 marks)
Ans: A unit vector in the direction of v is
u = 1/|v|* v
= 1 / √169 (3, 4, 12) = ( 3/13 , 4/13 , 12/13)
At the point (1, 2, −1 ):
∇ f ( x, y, z) = (2x + yz, −2y + xz, xy)
∇ f ( 1, 2, −1) = (0, −5, 2)
So, the directional derivative is:
Du f (1, 2, −1) = ∇ f (1, 2, −1 ).u = (0, −5 , 2) . (3/13 , 4/13 , 12/13) = 4/13
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