How to find the derivative of Ln (2x + 3) ^ 2?

How to find the derivative of Ln (2x + 3) ^ 2?

=2(2x+3)*2*(1/(2x+3)^2)
=4/(2x+3)
=1/(2x+3)²×2（2x+2)×2
=(8x + 8) / (2x + 3) & sup2; how to calculate?
If the polynomial P = A2 + 2B2 + 2A + 4B + 2008, then the minimum value of P is ()
A. 2005B. 2006C. 2007D. 2008
P = A2 + 2B2 + 2A + 4B + 2008, = (A2 + 2A + 1) + (2B2 + 4B + 2) + 2005, = (a + 1) 2 + 2 (B + 1) 2 + 2005, when (a + 1) 2 = 0, (B + 1) 2 = 0, P has the minimum value, and the minimum value is 2005
Find the derivative of the function f (x) = 1 / 2x at x = 2
Derivative = - 1 / 2x ^ 2 the derivative at 2 is - 1 / 8
f'(x)=(1/2)*x^(-1)=(1/2)*[-x^(-2)]=-1/2x²
∴f'(2)=-1/8
Derivative = - 1 / 2x ^ 2 the derivative at 2 is - 1 / 8
When a is what value, the polynomial a ^ 2 + B ^ 2 + 2a-4b + 16 has the minimum value? Find out the minimum value
^2 is the square
LIM (x - > 0) [f (2x) - f (0)] / x = 0.5, find the derivative of F (0)
lim[x→0] [f(2x)-f(0)]/x
=2lim[x→0] [f(2x)-f(0)]/(2x)
=2f '(0)
=1/2
So: F '(0) = 1 / 4
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When AB is a value, the polynomial a ^ 2 + B ^ 2-2a + 4B + 6 has a minimum value
A²+B²-2A+4B+6
=(A²-2A+1)+(B²+4B+4)+1
=(A-1)²+(B+2)²+1
∵(A-1)²≥0
(B-2)²≥0
Qi
A²+B²-2A+4B+6
=(A-1)²+(B+2)²+1
≥0+0+1
=1
In this case, A-1 = 0, B + 2 = 0
That is, a = 1, B = - 2
So when a = 1, B = - 2, the polynomial has a minimum value of 1
The function f (x) = - 2x3 + bx2 + CX (B, C ∈ R) defined on R is known. The function f (x) = f (x) - 3x2 is odd, and the function f (x) takes the extremum at x = - 1. (1) find the analytic expression of F (x); (2) discuss the monotonicity of F (x) in the interval [- 3, 3]
(1) ∵ function f (x) = f (x) - 3x2 is an odd function, ∵ f (- x) = - f (x), which is reduced to B = 3. ∵ function f (x) takes the extremum at x = - 1, ∵ f '(x) = 0. F (x) = - 2x3 + 3x2 + CX, f' (x) = - 6x2 + 6x + C ∵ f '(- 1) = - 6-6 + C = 0, C = 12
The minimum value of the algebraic expression 3A ^ 2 + 6 is
3a^2>=0
So the minimum value of 3A ^ 2 + 6 is 6
The definition domain of function f (x) is r, f (- 1) = 2. For any x ∈ R, f ′ (x) > 2, then the solution set of F (x) > 2x + 4 is ()
A. （-1，+∞）B. （-∞，-1）C. （2，+∞）D. （-∞，-2）
Let f (x) = f (x) - (2x + 4), then f (- 1) = f (- 1) - (- 2 + 4) = 2-2 = 0, and for any x ∈ R, f ′ (x) > 2, so f ′ (x) = f ′ (x) - 2 > 0, that is, f (x) increases monotonically on R, then the solution set of F (x) > 0 is (- 1, + ∞), that is, the solution set of F (x) > 2x + 4 is (- 1, + ∞)
When the minimum value of algebraic formula (a + 2) - 5 is obtained, the value of a is?
"The value of a when getting the minimum value is" do you have the minimum value in this algebraic expression
It's a quadratic function with the opening up, so it has a minimum. When a = - 2, it has a minimum of - 5