The known functions y1 sin 3x 4 , y2 3cos 3x 4 , let the functions y y1 y2 f x

f(x)=sin(3x-π/4)+√3cos(3x-π/4)=2[(1/2)sin(3x-π/4)+(3/2)cos(3x-π/4)]=2[cosπ/3sin(3x-π/4)+sinπ/3cos(3x-π/4)]=2sin(π/3+3x-π/4)

2sin(3x+π/12)

Let f(x)=2,-2, then 3x+ 12= 2+k x=( 2+k - 12) 3=k 3-5 36 (k=0,1,2...

This is the axis of symmetry equation.

Axis of symmetry center coordinates = (k 3-5 36,2) (k = 0,2,4...

or (k 3-5 36,-2) (k = 1,3,5....)

f(x)=sin(3x-π/4）+√3cos(3x-π/4）=2[1/2sin(3x-π/4)+√3/2cos(3x+π/4)]=2sin(3x-π/4+π/3)=2sin(3x+π/12)

Therefore, the axis of symmetry is x=5 36 k3, and k is an integer.

The center of symmetry is (- 36 k3) and k is an integer.

Summary. y’=6xcos（3x²+1）

Let +y=sin(3x +1), and + find +y

y’=6xcos（3x²+1）

Because y=sin(3x +1), y = cos(3x +1) (3x +1)'=cos(3x resells+1) 6x=6xcos(3x +1), so y'=6xcos(3x +1).

Because y=sin(3x +1), y = cos(3x +1) (3x +1)'=cos(3x resells+1) 6x=6xcos(3x +1), so y'=6xcos(3x +1).

3sinx+2e×-x+c

The rest of the problems need to be upgraded.

Summary. Hello dear customers, there are three questions in one round of middle and high school mathematics and two questions in one round of advanced mathematics, thank you for your understanding!

y=sin(4x +x-1), find y

Hello dear customers, there are three questions in one round of middle and high school mathematics and two questions in one round of advanced mathematics, thank you for your understanding!

To the process and the answer.

This question. y'The (8x 1)sin(4x +x-1) rigid process is the derivative of the composite function.

Change sin to cos

I just floated.

Can you write about it.

There is no pen around.

You split him into Sinu and Y 4x X-1

Then just derive each of these two functions.

Can you verbally describe the process?

I'll find you a pen to write later. Thank you.

Summary. Let y=sinx(x +1)y'=cosx(x +1)+2xsinx

Let y+=sinx(x +1) find y'

Let y=sinx(x +1)y'=cosx(x +1)+2xsinx, which can be seen as multiplying two functions

Can you be specific.

Leading and not leading + leading and not leading

Write about it. f（x）g（x）}'f（x）'g（x）=g（x）'f（x）

Substitute x=-1 into the function y=-2x+3

Get: y=-2 (-1)+3=5

Therefore, fill in 5

Summary. 6xcos (3x square).

y=sin(3x) then y =

6xcos (3x square).

y=xsin2x, then dy=

One less was written.

Plus it's fine.

Definite integral symbol.

cosx is a primitive function of f(x) in the interval i, then f(x)=sinxy=lnsinx, then y =

cosx/sinx

No, there is no smack. y=sin(3x) then y =

6xcos(3x²)

Solution: Because the table of this function is called the formula of the formula.

y (3xa1) (2x 1).

So, its derivative is .

y Tan Xiao [3 (2x 1) one 2 (3 x one 1)] 2x 1) Zheng Xinfeng 25 (2x 1) 2.

x +1 x -3x-3 kernels x +2 = 0

x²+2+1/x²-3x-3/x=0

x+1/x)²-3(x+1/x)=0

x+1/x)(x+1/x-3)=0

Because x+1 collapse x cannot be equal to zero, so.

x+1/x-3=0

x+1/x=3

x+1/x)³=27

x +3x+3 x+1 x number of burns = 27

x³+1/x³+3x+3/x=27

x³+1/x³+3(x+1/x)=27

x³+1/x³+3*3=27

x³+1/x³+9=27

x³+1/x³=18

y=-1/2x²-3x-2/5=(-x/2-5/2)(x+1)=-1/2(x+3)²+41/10

1) The coordinates of the intersection of the image posture of the function and the x-axis c(-5,0),b(-1,0), which is a parabola with a cluster opening towards the bottom, -5,1, known function y=-1 2x -3x-2 5

1) Find the coordinates of the intersection of the image of the function and the x-axis, and point out that when the value of x is taken, the value of the function is greater than zero.

2) Let the vertex of the function image be a, and its intersection with the x-axis is b,c, find the area of abc?