Finding the derivative of x power of y = cos root sign x plus 2

Finding the derivative of x power of y = cos root sign x plus 2

(1) Y = cos √ x + 2 ^ x → y '= - sin √ x · (√ x)' + 2 ^ xln2 = (- 1 / 2) (sin √ x) / √ x + 2 ^ xln2. (2) y = cos [(√ x) + 2 ^ x] → y '= - sin [(√ x) + 2 ^ x] · ((√ x) + 2 ^ x)' = - sin [((√ x) + 2 ^ x] · (1 / (2 √ x)) + 2 ^ xln2. I don't know which possibility
Given the function FX, if XY belongs to R, f (x + y) = FX + FY. If x ＞ 0, try to judge the monotonicity of FX in (0, positive infinity)
f(x+y)=f(x)+f(y)
Let x = y = 0
We get f (0 + 0) = f (0) + F (0)
So f (0) = 0
Let 0 "x 10"
So f (x2) = f [(x2-x1) + X1] = f (x2-x1) + F (x1)
f(x2)-f(x1)=f(x2-x1)
Because when x > 0, f (x)
Y = the second derivative of root x
Given that the function FX belongs to R for any x y, there is always FX + FY = f (x + y). If x > 0, FX
Let x = y = 0 2F (0) = f (0) f (0) = 0
Let y = - x, f (x) + F (- x) = f (0) = 0 - f (x) = f (- x) be an odd function
Find the monotone decreasing interval of y = 1 / 3x & # 179; - # 189; (a + A & # 178;) x & # 178; + A & # 179; X + A & # 178
y＝1／3x³-½﹙a＋a²﹚x²＋a³x＋a²
Derivation y '= x ^ 2 - (a + A ^ 2) x + A ^ 3 = (x-a) (x-a ^ 2)
If a = 0 or 1, then a = a ^ 2, y '≥ 0, y is an increasing function
If A0, then [a, a ^ 2] is a simple decreasing interval
If 0
It is known that the function FX is meaningful for x > 0 and satisfies F2 = 1 fxy = FX + FY. FX is an increasing function. If FX + F (X-2)) = 2 holds, then the value range of X is
f(2)=1
f(4)=f(2×2)=f(2)+f(2)=2
f(x)+f(x-2)=f[x(x-2)]
F (x) is an increasing function
x>0
x-2>0
x(x-2)≥4
The solution of inequality system is obtained
x≥1+√5
(x + 2Y & # 178;) - (x + y) (3x-y) where x = - 2, y = # 189; what's the problem?
The original formula = x & # 178; + 4xy + 4Y & # 178; - 3x & # 178; - 2XY + Y & # 178;
=-2x²+2xy+5y²
=-8-2+5/4
=-19/4
The original formula is reduced to x + 2Y & # 178; - (3x & # 178; + 2xy-y & # 178;) = - 2x & # 178; - 2XY + 3Y & # 178;
Substituting x = - 2, y = &#, the final answer is - 8 + 2 + 3 / 4 = - 5 and 1 / 4
If the function FX defined on R belongs to R for any x1.x2, f (x1 + x2) = FX1 + FX2 + 2 holds, and if x > 0, FX > - 2
1. Prove that GX = FX + 2 is an odd function
2. Prove that FX is an increasing function on R
3. If f (1) = - 1, solve the inequality f (log2m)
Let X1 = x2 = 0, so f (0 + 0) = f (0) + F (0) + 2, so f (0) = - 2
Let X1 = x, X2 = - x, so f (x-x) = f (x) + F (- x) + 2, so f (x) + F (- x) = f (0) - 2 = - 4
1.g(-x)=f(-x)+2,g(x)=f(x)+2
Because f (x) + F (- x) = - 4, f (- x) + 2 = - [f (x) + 2], i.e. g (- x) = - G (x), it is proved
2. Suppose X1 > 0, x2-x2
f(x1+x2)=f(x1)+f(x2)+2
Because X1 + x2 > 0, f (x1 + x2) > - 2
So f (x1) + F (x2) + 2 > - 2, f (x1) + 2 > - [f (x2) + 2] = f (- x2) + 2, then f (x1) > F (- x2) and G (x1) > G (- x2)
It is proved that when x > 0, G (x) is an increasing function. Because g (x) is an odd function, G (x) is an increasing function on R, and f (x) is also an increasing function on R
3.f(2)=f(1+1)=f(1)+f(1)+2=0,f(4)=2f(2)+2=2
So f (log2m)
How to judge monotonicity quickly according to derivative root
If the derivative is less than 0, the function decreases monotonically~
If the derivative is greater than 0, the function will increase monotonically!
For the function FX defined on the interval D, if it satisfies the condition of pair ᦉ 8704; x1, X2 ∈ D, and when X1 < X2, there is FX1 ≥ FX2, then the function FX is called a non increasing function
And f (0) = L, f (x) + F (l-x) = L, and if x ∈ [0,1 / 4], f (x) ≤ - 2x + 1 holds
①∀x∈[0,1],f（x）≥0；
② When x1, X2 ∈ [0,1] and x1 ≠ X2, f (x1) ≠ f (x);
③f(1/8)+f(5/11)+f(7/13)+f(7/8)=2;
④ When x ∈ [0,1 / 4], f (f (x)) ≤ f (x)
For detailed explanation
3. When x ∈ [0,1 / 4], f (x) ≤ - 2x + 1 is constant, then f (1 / 4) ≤ 1 / 2, because f (x) + F (l-x) = L, then f (1 / 2) = 1 / 2, f (1 / 4) = 1 / 2, f (3 / 4) = 1 / 2x ∈ (1 / 4,3 / 4) f (3 / 4) ≤ f (x) ≤ f (1 / 4), then f (x) is equal to 1 / 2, so f (5 / 11) = F