6.17) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text. Differentiation of composite functions6.17.Let z =f(x, y) and x = 0(t), y = y(1) where f, o, y are assumed differentiable. Provexp zedz dyдх dt ду dtdz%3DdtUsing the results of Problem 6.14, we havedz Axdz Aydz dx dz dydz= limAt0 AtAzAxAy+E2= limAt0 dx At+ɛ,dtdy AtAtAtdx dtdy dtAxdx Aydysince, as At → 0, we have Ax → 0, Ay → 0, ɛ, →0,->Atdt AtdtDifferentials6.14.Let f(x, y) have continuous first partial derivatives in a region R of the xy plane. Prove thatAf = f(x + Ax, y + Ay) -f(x, y) = fAx + f, Ay + 8, Ax + 8,Aywhere e, and e, approach zero as Ax and Ay approach zero.Applying the mean value theorem for functions of one variable (see Page 78), we haveAf = (f(x+ Ax, y + Ay) -f (x, y + Ay)} + {f(x, y + Ay) -f(x, y)}(1)= Arf,(x + 0, Ax, y + Ay) + Af,(x, y + 02 Ay)0 < 0, < 1,0< 0, <1Since, by hypothesis, f, and f, are continuous, it follows thatf,(x + 0, Ax, y + Ay) = f, (x, y) + 8,f,(x, y + 0,Ay) = f,(x, y) + 8,where d, → 0, 8, →0 as Ar →0 and Ay → 0.Thus, Af = f,Ax +fAy + 8,Ax + 8,Ay as required.Defining Ax dx, Ay = dy, we have Af f,dx +f, dy + 8, dx + 8,dy.We call df = f,dx + f, dy the differential of f (or z) or the principal part of Af (or Az).

Answered: Let z = f(x, y) and x = 0(1), y = y(1)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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6.17) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.

Differentiation of composite functions
6.17.
Let z =f(x, y) and x = 0(t), y = y(1) where f, o, y are assumed differentiable. Prove
xp ze
dz dy
дх dt ду dt
dz
%3D
dt
Using the results of Problem 6.14, we have
dz Ax
dz Ay
dz dx dz dy
dz
= lim
At0 At
Az
Ax
Ay
+E2
= lim
At0 dx At
+ɛ,
dt
dy At
At
At
dx dt
dy dt
Ax
dx Ay
dy
since, as At → 0, we have Ax → 0, Ay → 0, ɛ, →0,
->
At
dt At
dt
Differentials
6.14.
Let f(x, y) have continuous first partial derivatives in a region R of the xy plane. Prove that
Af = f(x + Ax, y + Ay) -f(x, y) = fAx + f, Ay + 8, Ax + 8,Ay
where e, and e, approach zero as Ax and Ay approach zero.
Applying the mean value theorem for functions of one variable (see Page 78), we have
Af = (f(x+ Ax, y + Ay) -f (x, y + Ay)} + {f(x, y + Ay) -f(x, y)}
(1)
= Arf,(x + 0, Ax, y + Ay) + Af,(x, y + 02 Ay)
0 < 0, < 1,0< 0, <1
Since, by hypothesis, f, and f, are continuous, it follows that
f,(x + 0, Ax, y + Ay) = f, (x, y) + 8,
f,(x, y + 0,Ay) = f,(x, y) + 8,
where d, → 0, 8, →0 as Ar →0 and Ay → 0.
Thus, Af = f,Ax +fAy + 8,Ax + 8,Ay as required.
Defining Ax dx, Ay = dy, we have Af f,dx +f, dy + 8, dx + 8,dy.
We call df = f,dx + f, dy the differential of f (or z) or the principal part of Af (or Az).

Expert Solution

Step 1

Given: $z = f (x, y) a n d x = ϕ (t), y = ψ (t)$

where $f, ϕ, ψ$ are assumed differentiable.

Since $f$ is differentiable

$\Rightarrow f (x, y)$ have continuous partial derivatives.

$\Rightarrow z$ have continuous partial derivatives.

Thus we can use theorem $6.14 .$

Let $f (x, y)$ have continuous first partial derivatives in a region $R$ of $x y -$ plane.

Then

$∆ f = f (x + ∆ x, y + ∆ y) - f (x, y) = f_{x} ∆ x + f_{y} ∆ y + δ_{1} ∆ x + δ_{2} ∆ y w h e r e δ_{1} \to 0, δ_{2} \to 0 a s ∆ x \to 0 a n d ∆ y \to 0$

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