How to find the derivative of Sint to the third power of Sint = x?

How to find the derivative of Sint to the third power of Sint = x?

I don't understand. Is the derivative cost a composite function

Let a and B be real numbers, and a + B = 3, then the maximum of 2 ^ A + 2 ^ B

If a + B = 3, then a = 3-b
SO 2 ^ A + 2 ^ B = 2 ^ (3-B) + 2 ^ B
Let 2 ^ B = x, then x > 0
Then 2 ^ A + 2 ^ B = 8 / x + x > = 2 √ ((8 / x) * x) = 4 √ 2, and when 8 / x = x, there is a maximum
So the maximum value of 2 ^ A + 2 ^ B = 4 √ 2

a. B is a real number a ^ 2 + B ^ 2 = a + B to find the maximum value of a + B

In summary, the maximum value of a + B is 2

If real numbers a and B satisfy | a + 1 | + | 2-A | = 5 - | B + 2 | - | B + 4 |, find the maximum value of a ^ 2 + B ^ 2

A:
|a+1|+|2-a|=5-|b+2|-|b+4|
|a+1|+|a-2|+|b+2|+|b+4|=5
It means the sum of the distances from point a to point - 1,2, and the sum of the distances from point B to point - 2, - 4 is 5
-1

It is known that a and B are positive real numbers, a + B + 1 / B + 4 / a = 10, then the maximum value of a + B is?

10＝a+b+1/b+4/a
＝a+b+1²/b+2²/a
≥(a+b)+(1+2)²/(a+b)
The results are as follows
(a+b)²-10(a+b)+9≤0
→1≤a+b≤9.
The maximum value is 9;
The minimum value is: 1

Given that real numbers a and B satisfy | a + 2 | + | 1-A | = 9 - | B-5 | - | 1 + B |, let the maximum value of a + B be m and the minimum value be n, then the value of M + n is m______ ．

The original formula is as follows: |a + 2 | + |a-1 + | B + 1 | + | B-5 | = 9, a is divided into three situations: a ≤ - 2, - 2 < a < a < 1, a ≥ 1, when a ≤ - 2, the |a + 2 + | B + 1 + | b-5-b-5-5 | 9 = 9, a is divided into three situations: a + 2, when a ≤ - 2, when a ≤ - 2, when a + 2 + 2 + + \124 + ＜ B ＜ 5, B ≥ 5 | B + 1 | + | B-5 | = - 2b + 4 or 6 When a ≤ - 2, B ≤ - 1, - 2a-1-2b + 4 = 9, a + B = - 3; when a ≤ - 2, - 1 ＜ B ＜ 5, - 2a-1 + 6 = 9, a = - 2, a + B maximum is less than 3, minimum is greater than - 3; when a ≤ - 2, B ≥ 5, - 2a-1 + 2b-4 = 9, A-B = - 7, a + B maximum is 3; when - 2 ＜ a ＜ 1, B ＜ - 1, 3-2b + 4 = 9, B = - 1, a + B maximum is less than 0 (5) when - 2 < a < 1, - 1 < B < 5, 3 + 6 = 9, maximum value of a + B is less than 6, minimum value is greater than - 3; (6) when - 2 < a < 1, B ≥ 5, 3 + 2b-4 = 9, maximum value of B = 5, maximum value of a + B is 6, minimum value is greater than 3; (7) when a ≥ 1, B ≤ - 1, 2A + 1-2b + 4 = 9, A-B = 2, maximum value of a + B is less than 0, minimum value is greater than - 2; (8) when a ≥ 1, B ≤ - 1, 2A + 1 + 6 = 9, a = 32, minimum value is greater than - 2 +When a ≥ 1, B ≤ - 1, 2A + 1 + 2b-4 = 9, ｜ a + B = 6; finally, it is concluded that the maximum value of a + B can be 6, and the minimum value is - 3, so m + n = 3

Given positive real numbers a and B, satisfying a + B = 1, find the maximum or minimum of [A / (1 + b)] + [B / (1 + a)] (online, etc.)
Such as the title

If positive real numbers a and B satisfy a + B = 1, find the maximum or minimum of B / (1 + a) + A / (1 + b)

B (1 + a) + A / (1 + b) = (1-A) / (1 + a) + (1-B) / (1 + b) = - 1 + 2 / (1 + a) - 1 + 2 / (1 + b) = - 2 + 2 [1 / (1 + a) + 1 / (1 + b)] that is to find the maximum value of 1 / (1 + a) + 1 / (1 + a) + 1 / (1 + b) = (1 + A + 1 + b) / (1 + a) (1 + B) = 3 / (1 + a) (1 + 1-A) = 3 / (- A ^ 2 + A + 2) = 3 / [- (A-1 / 2) ^ 2 + 9 / 4] so a = 1 / 2 b = 1

Given that the maximum value of y = a + bcos3x is 3 / 2 and the minimum value is - 1 / 2, then the value of real number B is

y=a+bcos3x
b> 0 hour
Maximum = a + B = 3 / 2
Minimum = A-B = - 1 / 2
Solution
2b=2
b=1
b

If a and B belong to positive real numbers, 2A + 3B = 4., then the maximum value of ab

It can be solved by means inequality
2A + 3b ≥ 2 √ (2a) · √ (3b), and 2A + 3B = 4, so 2 √ (2a) · √ (3b) ≤ 4, sort out √ (6ab) ≤ 2, square, ab ≤ 2 / 3, when 2A = 3b, the equal sign holds, at this time a = 1, B = 2 / 3, the maximum value of AB is 2 / 3