Basic Concept in The directional derivative of a function f(x,y) along a vector V = (V₁, V₂) is the function Dut defined by the (Dt) (x,y) = lim 4-30 Notation CUS Du f(x) = f(x) - V = v. af(x) dx. f(x+v₁h₂y+v₂h)-f(x,y) h Now by definition, Du f (x₁y) = bem f(x+ V₁h, Y + V₂h)-f(x,y) 410 Using chain Rule, we can Duf(x, y) = f(x) dx N+ write. f (x, y) ay -NSL let We have, f (x,y) = lim f(x=Nih, YEN₂h) - f (x,y) 830 We can write, v₁ + f(y) == √2+(2₁9) Əz After comparing & Ⓡ еле деву Duf h af (x₁y) v ₂ Dy f Proved Question 224 using, Dy (et) (x,y) = Limf(x+hu;y+hus)-8-f(x,y) h show that, Q₂ (cf) = c Dyf. ⇒ lim [f(x+v₁h, y + V₂h)-f(xy)] erto h 140 = C Hence, Using Product sucts D₂ (cf)(x,y)== 12+ (x,y) v₁ +t 2x for differentiation 2t (x,y) v₂ of [3+ (xy³v₁ + 3+ (x,y) v₂] D(c)(x,y) = D f (rey) Dy (cf) = c Duf 11.
Basic Concept in The directional derivative of a function f(x,y) along a vector V = (V₁, V₂) is the function Dut defined by the (Dt) (x,y) = lim 4-30 Notation CUS Du f(x) = f(x) - V = v. af(x) dx. f(x+v₁h₂y+v₂h)-f(x,y) h Now by definition, Du f (x₁y) = bem f(x+ V₁h, Y + V₂h)-f(x,y) 410 Using chain Rule, we can Duf(x, y) = f(x) dx N+ write. f (x, y) ay -NSL let We have, f (x,y) = lim f(x=Nih, YEN₂h) - f (x,y) 830 We can write, v₁ + f(y) == √2+(2₁9) Əz After comparing & Ⓡ еле деву Duf h af (x₁y) v ₂ Dy f Proved Question 224 using, Dy (et) (x,y) = Limf(x+hu;y+hus)-8-f(x,y) h show that, Q₂ (cf) = c Dyf. ⇒ lim [f(x+v₁h, y + V₂h)-f(xy)] erto h 140 = C Hence, Using Product sucts D₂ (cf)(x,y)== 12+ (x,y) v₁ +t 2x for differentiation 2t (x,y) v₂ of [3+ (xy³v₁ + 3+ (x,y) v₂] D(c)(x,y) = D f (rey) Dy (cf) = c Duf 11.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Basic Concept or
The directional derivative of a
function f(x,y) along
V = (V₁, V₂) is the function Dut
defined by the,
(D₂+) (x₁y) = lim f(x+v₁h;y+v₂h) = f(x,y)
f)
4-0
h
Notation as
Du f(x)
Dvf(x,y)
= f(x) • V = V.
Using chain Rule, we can
a vector
Now by definition,
Du f (x₁y) = bem f(x+ V₁h, Y+V₂h)-f(x,y)
410
If (114)
In
V₁ +
h
write.
Jf(x)
2x.
of (x, y)
ay
-№₂
*
let
We have,
<-V₁₂ -√₂>]
D_vif (x, y) = lim
8-30
f(x=N₁h, y=N₂h) -f (my)
We can write,
Du f(x₁y) = [2 f(x,y).
2x
v₁ +
After comparing ***
сле gets
h
af ( x ₁ y) v ₂
ду
Duf'= - Duf
[V₂]
Proved
Question 24"
Using,
Dv (et) (x,y) = lim
ht0
D₁ (cf) (x₁y) = C
⇒ lim [cf(x+v₁h, y + V₂h) - Of(xy)]
hto
h
D(cf)(x4)
Show that,
Using Product rule for differentiation
af (x,y) vz
ду
Hence,
Q₁₂ (cf) = c Dyf.
=
f (x +hv, y + hv₂) - Of(24)
h
² [
of (x,y) v₁ +0
2x
of
of
[3+ (x,y), + 3+ (x,y) √₂]
Ix
ду
Du f (x,y)
Dv (cf) = c Duf
"1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05acc741-d43b-42c1-835f-678a6d9974b9%2F1d36bf28-7018-4d1a-9231-2b78c18de340%2Ffczo9lp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Basic Concept or
The directional derivative of a
function f(x,y) along
V = (V₁, V₂) is the function Dut
defined by the,
(D₂+) (x₁y) = lim f(x+v₁h;y+v₂h) = f(x,y)
f)
4-0
h
Notation as
Du f(x)
Dvf(x,y)
= f(x) • V = V.
Using chain Rule, we can
a vector
Now by definition,
Du f (x₁y) = bem f(x+ V₁h, Y+V₂h)-f(x,y)
410
If (114)
In
V₁ +
h
write.
Jf(x)
2x.
of (x, y)
ay
-№₂
*
let
We have,
<-V₁₂ -√₂>]
D_vif (x, y) = lim
8-30
f(x=N₁h, y=N₂h) -f (my)
We can write,
Du f(x₁y) = [2 f(x,y).
2x
v₁ +
After comparing ***
сле gets
h
af ( x ₁ y) v ₂
ду
Duf'= - Duf
[V₂]
Proved
Question 24"
Using,
Dv (et) (x,y) = lim
ht0
D₁ (cf) (x₁y) = C
⇒ lim [cf(x+v₁h, y + V₂h) - Of(xy)]
hto
h
D(cf)(x4)
Show that,
Using Product rule for differentiation
af (x,y) vz
ду
Hence,
Q₁₂ (cf) = c Dyf.
=
f (x +hv, y + hv₂) - Of(24)
h
² [
of (x,y) v₁ +0
2x
of
of
[3+ (x,y), + 3+ (x,y) √₂]
Ix
ду
Du f (x,y)
Dv (cf) = c Duf
"1
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