Let the function f (x) = cos (2x + three branches π) + sin square x find the maximum and minimum period

Let the function f (x) = cos (2x + three branches π) + sin square x find the maximum and minimum period

F (x) = cos2x * cos row / 3-sin2x * sin row / 3 + (1-cos2x) / 2
f(x)=cos2x/2-(√3/2)sin2x-(1/2)cos2x+1/2
f(x)=-(√3/2)sin2x+1/2
The maximum value is √ 3 / 2 + 1 / 2 = (√ 3 + 1) / 2
Minimum cycle is 2 rows / 2 = rows

Known function f (x) = sin (π)- ω x)cos ω Square of X + cos ω x( ω> The minimum positive period of 0) is π 1. Seek ω The abscissa of a point on the image of function y = f (x) is shortened to the original 1 / 2, and the ordinate remains unchanged. The image of function y = g (x) is obtained, and the minimum value of function y = g (x) in the interval [0, π / 16] is calculated

f(x)=sin(π- ω x)cos ω Square of X + cos ω X = sinwxcoswx + (cos2wx + 1) / 2 = 1 / 2sin2wx + (cos2wx + 1) / 2 = root 2 / 2Sin (2wx + π / 4) + 1 / 2T = 2 π / 2W = π w = 1 shorten the abscissa of the last point on the image of function y = f (x) to the original 1 / 2, and shorten the unchanged period of ordinate to half of the original

Function y = (SIN) ² x) ²＋ (cos ² x) ² Minimum positive period of

y=(sin2x)2＋(cos2x)2
Formula = 1-2sin2xcos2x
=1-（sin2x）2/2
Find the period of sin2x absolute value
It should be π / 2

Derivative of Ln (2x ^ 2 + 3x + 1) I can't understand the examples in the textbook. Solve them. Help me, please!

I don't understand the examples in the textbook. I don't understand the derivation of composite functions
LNX derivation = 1 / X
Ln (f (x)) derivative = 1 / F (x) times the reciprocal of F (x), OK, the problem is solved

Ln (2x ^ 2-3x + 1) second derivative

The first derivative is (4x-3) / (2x ^ 2-3x + 1)
The second derivative is [4 (2x ^ 2-3x + 1) - (4x-3) ²]/ (2x^2-3x+1) ²

Find the derivative of y = (2x ^ 2; - 1) (3x + 1)

y=(2x^2;-1)`(3x+1)+(2x^2;-1)(3x+1)`=(4x)(3x+1)+(2x^2;-1)*3=12x^2+4x+6x^2-3=18x^2+4x-3