Chap 2:
1. Partial Derivatives
Function z = f(x,y)
+ First-order derivatives : 𝑧′𝑥 𝑎𝑛𝑑 𝑧′𝑦
Ex : Let 𝑓(𝑥, 𝑦) = 2𝑥 2 + 9𝑥𝑦 3 + 8𝑦 + 5
We have first-order derivative : 𝑓𝑥 (𝑥; 𝑦) = 2.2𝑥 + 9. 𝑦 3 = 4𝑥 + 9𝑦 3
𝑓𝑦 (𝑥, 𝑦) = 9. 𝑥. 3𝑦 2 + 8 = 27𝑥𝑦 2 + 8
Meaning: If 𝒇𝒙 (𝒙; 𝒚) = 𝜶 then keeping y unchanged (𝒚 = 𝒚𝟎 ), increasing x from 𝒙𝟎 𝒕𝒐 𝒙𝟎 + 𝟏 will
increase z by 𝜶
+Second-order derivatives (4) 𝑧′𝑥𝑦 , 𝑧′𝑦𝑥 , 𝑧′𝑥𝑥 𝑎𝑛𝑑 𝑧′𝑦𝑦
Ex : Above function  𝑧′𝑥𝑦 = 9.3𝑦 2 = 27𝑦 2 𝑧′𝑥𝑥 = 4
𝑧′𝑦𝑥 = 27𝑦 2 𝑧′𝑦𝑦 = 27𝑥. 2𝑦 = 54𝑥𝑦
2. Maxima and minima
Test for relative extrema:
B1: Find all derivatives : 𝑓𝑥 , 𝑓𝑦 , 𝑓𝑥𝑥 , 𝑓𝑦𝑦 , 𝑓𝑥𝑦
B2: Find critical points: from equation system 𝑓𝑥 = 0 𝑎𝑛𝑑 𝑓𝑦 = 0
B3: Calculate D  f xx (a, b)  f yy (a, b)   f xy (a, b) 
2
B4: Check




If
If
If
If
D  0 and f xx (a, b)  0 , then f (a, b) is a relative maximum.
D  0 and f xx (a, b)  0 , then f (a, b) is a relative minimum.
D  0 , the f(x,y) has a saddle point at (a,b).
D  0 then the test fails.
B5: Conclusion
Ex: Cho hàm f ( x, y )  2 x 2  3 y 2  4 x  3 y  5
-
Tìm tất cả đạo hàm
𝑓𝑥 (𝑥, 𝑦) = −4𝑥 − 4 𝑓𝑦 (𝑥, 𝑦) = −6𝑥 + 3
f xx ( x, y)  4, f yy ( x, y)  6, f xy ( x, y)  f yx ( x, y)  0
-
Tìm Critical points
f x ( x, y )  4 x  4
0  4 x  4 and
x  1
f y ( x, y)  6 y  3
0  6 y  3
1
y
2
The critical point for the given function is at  1, 1  .

-
Evaluate
2
D  f xx ( a, b)  f yy ( a, b)   f xy ( a, b) 
-
2
 ( 4)  ( 6)  0
 24
Since D > 0 and f xx (a, b)  4  0 we conclude that a relative maximum exists at
 1, 12 , f (1, 12)    1, 12 ,7.75 .
3. Larange multiplier
Find maximum and minimum subject to…. (g(x)=0)
B1: Write constraint function as g(x)=0 (the right side must be 0)
B2: Write Lagrange function
𝑭(𝒙, 𝒚, 𝝀) = 𝒇(𝒙, 𝒚) − 𝝀𝒈(𝒙, 𝒚) (𝜆: Lagrange multiplier)
B3: Find 𝑓𝑥 (𝑥, 𝑦, λ), 𝑓𝑦 (𝑥, 𝑦, λ) 𝑎𝑛𝑑 𝑓λ (𝑥, 𝑦, λ)
𝑓𝑥 (𝑥, 𝑦, λ) = 0
B4: Solve equation system {𝑓𝑦 (𝑥, 𝑦, λ) = 0
𝑓λ (𝑥, 𝑦, λ) = 0
 (x,y)
B5: Substitute (x,y) above to evaluate Max/Min
B6: Conclusion
4. Double integrals
A double integral is something of the form
∬ 𝑓(𝑥, 𝑦)𝑑𝑥. 𝑑𝑦
𝑅
where R is called the region of integration and is a region in the (x, y) plane. The double integral gives
us the volume under the surface z = f(x, y), just as a single integral gives the area under a curve
 Áp dụng tính Thể tích
Problem example:
Note: Calculate from inside to outside
Problems: Interchanging Limits of Limitation (Apply if not first integrate the function with respect
to x)