F (x) = vector a * vector B, known vector a = (2 √ 3sinx, cosx), vector b = (cosx, - 2cosx) 1) Find the analytic expression of F (x) and change it into the form of asinx (ω x + φ) + K 2) If G (x) = PF (x) + Q, there is a maximum value of 4 and a minimum value of - 2 on X ∈ [0, π / 2], find the value of P and Q

F (x) = vector a * vector B, known vector a = (2 √ 3sinx, cosx), vector b = (cosx, - 2cosx) 1) Find the analytic expression of F (x) and change it into the form of asinx (ω x + φ) + K 2) If G (x) = PF (x) + Q, there is a maximum value of 4 and a minimum value of - 2 on X ∈ [0, π / 2], find the value of P and Q

(1) F (x) = vector a * vector b = (2 √ 3sinx, cosx) · (cosx, - 2cosx) = 2 √ 3sinxcosx-2cos ^ 2x = √ 3sin2x-cos2x + 1 = 2Sin (2x - π / 6) + 1 (2) ∵ 0 ≤ x ≤ π / 2, ∵ - π / 6 ≤ 2x - π / 6 ≤ 5 π / 6 ∵ 2x - π / 6 = π / 2, there is a maximum of 4, ∵ 3P + q = 4 ① 2x - π / 6 = - π / 6

D ^ 2Y / DX ^ 2 = (dy / DX) '× (dy / DX). In addition, please explain the meaning of DX and dy. do DX and Dy refer to the formulas of x =... And y =... (such as parametric equation)? Or the derivatives of X and y... I'm confused here

No, d ^ 2Y / DX ^ 2 is a second derivative, not a power. Dy / DX is a first derivative, equivalent to f '

Can d ^ 2Y / DX ^ 2 be equal to (dy / DX) and then divided by DX
RT

D & sup2; Y / DX & sup2; denotes the second derivative of function y (x)
That is, D & sup2; Y / DX & sup2; = D (dy / DX) / DX. The understanding of the building owner only needs to differentiate (dy / DX)
Derivative can also be understood as division, that is, the differential of Y divided by the differential of X
The knowledge of higher mathematics is constructed by elementary mathematics, not created out of thin air
The above is my personal understanding,

What's the difference between D (DX) and D (Dy)

Guo Dunyong replied:
Generally, in a function, X is the independent variable and Y is the dependent variable. DX is the differential of the independent variable x and Dy is the differential of the dependent variable y. D (DX) is the second order differential of the independent variable x and D (Dy) is the second order differential of the dependent variable y
(the second derivative y ″ = D & # 178; Y / DX & # 178; is not expressed as y ″ = D & # 178; Y / D & # 178; x)

Y = 2 (x ^ 3) + 5x, when x = 3, what is dy / DX

dy/dx=6x^2+5 so when x=3 dy/dx=59

Ycos (Y / x) = ((x ^ 2 / y) * sin (Y / x) + xcos (Y / x)) dy / DX what method should I use to solve this problem
Let's talk about my steps... DZ / DX = (DZ / dy) * (dy / DX) = (1 / x) * dy / DX
(sinz/(z^2*cosz)+1/z)dz=dx/x
And then it's part by part Integration... But it's getting more and more complicated, and it can't be solved
Maybe the equation is not clear: ycos (Y / x) = ((x ^ 2) / y) * sin (Y / x) + xcos (Y / x)) dy / DX
Finally, find the expression of Y with respect to X

Find the general solution of differential equation ycos (Y / x) = [(X & # / y) sin (Y / x) + xcos (Y / x)] dy / DX
Let u = Y / x, then y = UX, dy / DX = u + XDU / DX
uxcosu=[(x/u)sinu+xcosu](u+xdu/dx)=xsinu+uxcosu+x²[(1/u)sinu+cosu](du/dx)
It is reduced to: xsinu + X & # 178; [(1 / U) sinu + COSU] (DU / DX) = 0, that is, sinu + x [(1 / U) sinu + COSU] (DU / DX) = 0
Separation of variables: DX / x + [(sinu + ucosu) / usinu] Du = 0
That is: DX / x + [(1 / U) + (COSU / sinu)] Du = 0
Integral: LNX + LNU + lnsinu = LNC
So there is xusinu = C, and substituting u = Y / x, the general solution is: ysin (Y / x) = C

Dy / DX = 1 / x + y, please use the first-order linear equation solution, do not use the substitution x + y = t to find that I will

dy/dx=1/(x+y)
dx/dy=x+y
This is a first order differential equation about X
It is easy to get the solution as follows
x=Ce^y+y-1

The - 2x power cos3x of y = e,

dy/dx=(e^-2x)'*cos3x+e^(-2x)*(cos3x)'
=e^(-2x)*(-2x)'*cos3x+e^(-2x)*(-sin3x)*(3x)'
=-2e^(-2x)cos3x-3e^(-2x)*sin3x

Let {x = Sint, y = cos2t, then dy / dx|t = π / 4=

dx/dt=cost=√2/2
dy/dt=-2sin2t=-2
So dy / dx|t = π / 4 = dy / dt △ DX / dt = - 2 √ 2

X = 2T ^ 2 + 1, y = Sint, how much is dy / DX?
X is the square of 2T plus 1,

dy/dt=cost
dx/dt=4t
dy/dx=cost/4t