F (x) = x + A. what about a when finding the derivative of FX

F (x) = x + A. what about a when finding the derivative of FX

Is 0
2a**2-8ab+17b**2-16a-4b+68
a=8 b=2
(a+b)^b=8^2=100
One hundred
Given that f (x) = ln (2x + 1), if the derivative of F (x) + F (x) = a has a solution, find the value range of A
Let g (x) = ln (2x + 1) + 2 / (2x + 1) = a, let g (x) = ln (2x + 1) + 2 / (2x + 1). If G (x) = ln (2x + 1) + 2 / (2x + 1), the derivative of G (x) is increasing at (1 / 2, + OO), and f (x) = ln (2x + 1). If the derivative of F (x) + F (x) = a has a solution, the solution is a > = LN2 + 1
Known P = 2A ^ 2-8ab + 17b ^ 2-16a-4b + 2069 math problem
It is known that P = 2A ^ 2-8ab + 17b ^ 2-16a-4b + 2069 is the minimum value of P, which means that when p is the minimum value, the value of AB is the minimum value
Find the derivative y = f (e ^ x) e ^ [f (x)]
y′=f′(e^x)（e^x）（e^(f(x))）+f(e^x)(e^(f(x)))(f′(x))
Let f (x) be an odd function defined on R, and if x ∈ 0, + ∞, f (x) = x (1 + 3 √ x), find the analytic expression of F (x)
Give the whole process
X∈-∞,0 -x∈0,+∞
Then - f (x) = f (- x) = - x (1 + 3 √ - x)
Then f (x) = x (1 + 3 √ - x)
Repartition function
Find the second derivative y = e ^ (x ^ 2) f (x ^ (e ^ 2))
The combination of the derivative rule of function product and the derivative rule of compound function has nothing but trouble in writing
 
Let f (x) be an odd function on R. when x ∈ (0, + ∞), f (x) = x (1 + 3 √ x), find the expression on f (x) Zai (- ∞, 0)
Where 3 √ x is the root of the third power
Because f (x) is an odd function,
When x belongs to (- infinity, 0), f (x) = - f (- x)
This is the expression that - x belongs to (0, + ∞), substituting (- x) into F when x is greater than 0
f(x)=-f(-x)=-((-x)*(1+3√(-x)))=x(1+3√x)
f(x)=（-x）*(1+（-x）^(1/3)) x∈（-∞，0）
^Denotes the power
- -|
Find the derivative of F (x) = (x ^ 2) y e ^ (2Y - x ^ 3) &# 160
Find the derivative of the square of x times the exponent of Y times e (2Y minus the third power of x)
Y is a constant
f'(x)= 2xy e ^(2y -x^3)+ (x^2 ) y e ^(2y -x^3)*(-3x^2)
=2xy e ^(2y -x^3)-3（x^4 ) y e ^(2y -x^3)
＝(2-3x^3)xye^(2y-x^3)
Let f (x) be an odd function defined on R, and if x ∈ [0, + ∞), f (x) = x (1 + x ^ 3), then if x ∈ (- ∞, 0], f (x)=
f（x）=x-x^4
When x ∈ (- ∞, 0), - x ∈ [0, + ∞), then f (x) = - f (- x) = - [(- x) (1 + (- x) ^ 3)] = x (1-x ^ 3). Question: is it not x (x ^ 3-1)?