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If x = 12 / π, how much is Cos4 to SiN4
Cos4th power - sin4th power
=(cos^2x-sin^2x)(cos^2x+sin^2x)
=cos2x
=cos(2*pai/12)
=Root 3 / 2
4-channel height number derivation solution: y = arccos (1-2x) y = lncot (x / 2) y = e to the negative third power of X × Sin3x y = square of COS x times cos (square of x)
Using the derivative method of composite function, it is very simple. 1, y '= - 1 / √ [1 - (1-2x) ^ 2] * (- 2) = 2 / √ (4x-4x ^ 2) = 1 / √ (x-x ^ 2) 2, y' = 1 / cot (x / 2) * [- CSC ^ 2 (x / 2)] * 1 / 2 = sin (x / 2) / cos (x / 2) * [- 1 / sin ^ 2 (x / 2)] * 1 / 2 = - 1 / [2Sin (x / 2) cos (X / 2)] = 1 / sin (x / 2) = CSC (x / 2) 3
On differential Suppose the second derivative of F (x) exists Prove that the second derivative of F (x) is equal to When x approaches 0 Limit of [f (x + H) - f (X-H) - 2F (x)] / h ^ 2
It should be that h tends to 0, and there should be a plus sign between F (x + H) and f (X-H)
The second derivative of F (x) exists, so it is second-order differentiable in the domain
For LIM [f (x + H) + F (X-H) - 2F (x)] / h ^ 2, use lobida's law to find the derivative of H
=[f'(x+h)-f'(x-h)]/2h
Derivative again
=[f''(x+h)+f''(x-h)]/2
=2f''(x)/2
=f''(x)
Find the general solution of differential equation with a higher number, YY '' - (y ') ^ 2 + y' = 0
Let P = y '
Then y "= DP / DX = DP / dy * dy / DX = PDP / dy
Substitute into the original equation: ypdp / DY-P ^ 2 + P = 0
Get: P = 0 or ydp / DY-P + 1 = 0
P = 0: dy / DX = 0, i.e. y = C
Ydp / DY-P + 1 = 0, DP / (p-1) = dy / y, ln (p-1) = LNY + C1, P-1 = cy
Get: dy / DX = CY + 1,
Get: dy / (CY + 1) = CX,
Ln (CY + 1) = CX ^ 2 / 2 + C2
cy+1=e^(cx^2/2+c2)
y=[e^(cx^2/2+c2)-1]/c
YY '' - (y ') ^ 2-y' = 0 to find the general differential solution
Let y '= P, then y "= DP / DX = DP / dy · dy / DX = P · DP / Dy, so the original equation is changed into YP · DP / DY-P ^ 2-P = 0, that is, P [y · DP / dy - (P + 1)] = 0, so p = 0 or Y · DP / dy = P + 1. For P = 0, y = (C1) can be solved; for y · DP / dy = P + 1, Y / py = (P + 1) / DP = P / DP + 1 / DP, that is, py / y = DP / (P + 1), LNY = ln (
Higher number problems. Finding the general solution of differential equations (2) x + YY '= 0 (4)
x+yy '=0
y·dy/dx=-x
y·dy=-x·dx
Points at both ends:
∫y·dy=∫-x·dx
y ²/ 2=-x ²/ 2+C1
Y ²+ x ²= 2C1
Let C = 2C1
Get y ²+ x ²= C
So the general solution of the differential equation is: y ²+ x ²= C
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