if w=f(u,v) where u=X+y, v=x-y show DW/DX +DW/Dy =2dw/du solve
Answers
Answer:
11. Differentials and the chain rule
Let w = f(x, y, z) be a function of three variables. Introduce a new
object, called the total differential.
df = fx dx + fy dy + fz dz.
Formally behaves similarly to how ∆f behaves,
∆f ≈ fx ∆x + fy ∆y + fz ∆z.
However it is a new object (it is not the same as a small change in f as
the book would claim), with its own rules of manipulation. For us, the
main use of the total differential will be to understand the chain rule.
Suppose that x, y and z are functions of one variable t. Then w =
f(x, y, z) becomes a function of t. Divide the equation above to get the
derivative of f,
df
dt = fx
dx
dt + fy
dy
dt + fz
dz
dt.
This is an instance of the chain rule.
Example 11.1. Let f(x, y, z) = xyz+z
2
. Suppose that x = t
2
, y = 3/t
and z = sin t.
Then
fx = yz fy = xz and fz = 2z,
so that
dw
dt = 2yzt −
3xz
t
2
+ (xy + 2z) cost = 3 sin t + (3t + 2 sin t) cost.
On the other hand, if we substitute for x, y and z, we get
w = 3tsin t + sin2
t,
and we can calculate directly,
dw
dt = 3 sin t + 3t cost + 2 sin t cost.
There are two ways to see that the chain rule is correct.
dx = x
0
(t) dt dy = y
0
(t) dt and dz = z
0
(t) dt.
Substituting we get
dw = fx dx + fy dy + fz dz
= fxx
0
(t) dt + fyy
0
(t) dt + fzz
0
(t) dt,
and dividing by dt gives us the chain rule.
More rigorously, start with the approximation formula,
∆w ≈ fx ∆x + fy ∆y + fz ∆z,
1
divide both sides by ∆t and take the limit as ∆t → 0.
One can use the chain rule to justify some of the well-known formulae
for differentiation.
Let f(u, v) = uv. Suppose that u = u(t) and v = v(t) are both
functions of t. Then
d(uv)
dt = fu
du
dt + fv
dv
dt = vu0 + uv0
,
which is the product rule. Similarly if f = u/v, then
d(u/v)
dt = fu
du
dt + fv
dv
dt =
1
v
u
0 −
u
v
2
v
0 =
u
0
v − v
0u
v
2
,
which is the quotient rule.
Now suppose that w = f(x, y) and x = x(u, v) and y = y(u, v).
Then
dw = fx dx + fy dy
= fx(xu du + xv dv) + fy(yu du + yv dv)
= (fxxu + fyyu) du + (fxxv + fyyv) dv.
= fu du + fv dv.
If we write this out in long form, we have
∂f
∂u =
∂f
∂x
∂x
∂u +
∂f
∂y
∂y
∂u and ∂f
∂v =
∂f
∂x
∂x
∂v +
∂f
∂y
∂y
∂v .
Example 11.2. Suppose that w = f(x, y) and we change from Carte-
sian to polar coordinates,
x = r cos θ
y = r sin θ.
We have
∂x
∂r = cos θ
∂x
∂θ = −r sin θ
∂y
∂r = sin θ
∂x
∂θ = r cos θ.
So
fr = cos θfx + sin θfy
fθ = −r sin θfx + r cos θfy.
2
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