Partial derivatives
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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Question
6.8) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
![Partial derivatives
6.8.
If f (x, y) = 2x? – xy + y², find (a) d fl d x and (b) d fld y at (xo, Yo) directly from the definition.
af
(a)
ax
f(x, + h, y,) – f(x,, Yo)
= f,(x,, Yo) = lim
h
[2(x, + h)' – (x, + h)y, + y°]=[2x, – x,Yo + y°]
= lim
h
4hx, + 2h? – hy
lim
-
3 lim(4x, + 2h - y,) %3D 4x, — у
h→0
h
h0
af
(b)
ду
f(Xp, Yo + k) – f(xp, Yo)
= f,(x, Yo) = lim
k
[2x, – x,(Yo + k) + (y, + k)']– [2x, – x,Vo + y°]
= lim
k
-kx, + 2ky, + k²
= lim
lim(-x, + 2y, + k)= – x, + 2y%
k
Since the limits exist for all points (Xp, Yo), we can write f,(x, y) =f;= 4x – y, f, (x, y) =fy=-x+2y, which
are themselves functions of x and y.
Note that formally f,(xo, Yo) is obtained from f(x, y) by differentiating with respect to x, keeping y constant
and then putting x = Xp, y = Yo- Similarly, f,(Xo, Yo) is obtained by differentiating f with respect to y, keeping x
constant. This procedure, while often lucrative in practice, need not always yield correct results (see Problem
6.9). It will work if the partial derivatives are continuous.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2F90319ba9-e962-4356-b6a4-e8db125b1591%2Fwwmrg3m_processed.png&w=3840&q=75)
Transcribed Image Text:Partial derivatives
6.8.
If f (x, y) = 2x? – xy + y², find (a) d fl d x and (b) d fld y at (xo, Yo) directly from the definition.
af
(a)
ax
f(x, + h, y,) – f(x,, Yo)
= f,(x,, Yo) = lim
h
[2(x, + h)' – (x, + h)y, + y°]=[2x, – x,Yo + y°]
= lim
h
4hx, + 2h? – hy
lim
-
3 lim(4x, + 2h - y,) %3D 4x, — у
h→0
h
h0
af
(b)
ду
f(Xp, Yo + k) – f(xp, Yo)
= f,(x, Yo) = lim
k
[2x, – x,(Yo + k) + (y, + k)']– [2x, – x,Vo + y°]
= lim
k
-kx, + 2ky, + k²
= lim
lim(-x, + 2y, + k)= – x, + 2y%
k
Since the limits exist for all points (Xp, Yo), we can write f,(x, y) =f;= 4x – y, f, (x, y) =fy=-x+2y, which
are themselves functions of x and y.
Note that formally f,(xo, Yo) is obtained from f(x, y) by differentiating with respect to x, keeping y constant
and then putting x = Xp, y = Yo- Similarly, f,(Xo, Yo) is obtained by differentiating f with respect to y, keeping x
constant. This procedure, while often lucrative in practice, need not always yield correct results (see Problem
6.9). It will work if the partial derivatives are continuous.
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