A square cloth has a side length of 0.6 meters. What are the perimeter and area of the square cloth?

A square cloth has a side length of 0.6 meters. What are the perimeter and area of the square cloth?

The perimeter of this square cloth is 0.6 * 4 = 2.4 meters
The area is 0.6 * 0.6 = 0.36 square meters

Find the derivative of the function determined by the following parametric equation
{x=3e^（-t） ;
y=2e^t
An advanced number problem

E ^ t = 3 / x from x = 3E ^ (- t)
Substituting y = 2E ^ t = 2 * 3 / x = 6 / X
So y '= - 6 / x ^ 2

Monotone interval and range of y = 2Sin (- 1 / 2x - π / 6) on [0,2 π]
I'll raise the reward and finish in three hours

y=2sin(-1/2x-π/6)
=-2sin(1/2x+π/6)
The increasing interval of function sin (1 / 2x + π / 6) is
Decreasing interval of y = - 2Sin (1 / 2x + π / 6)
The decreasing interval of function sin (1 / 2x + π / 6) is
The increasing interval of y = - 2Sin (1 / 2x + π / 6)
From 2K π - π / 2 ≤ 1 / 2x + π / 6 ≤ 2K π + π / 2
We obtain that 4K π - 4 π / 3 ≤ x ≤ 4K π + 2 π / 3, K ∈ Z
Taking k = 0, the monotone decreasing interval on [0,2 π] is obtained
[0,2π/3],
From 2K π + π / 2 ≤ 1 / 2x + π / 6 ≤ 2K π + 3 π / 2
We obtain that 4K π + 2 π / 3 ≤ x ≤ 4K π + 8 π / 3, K ∈ Z
Taking k = 0, the monotone increasing interval on [0,2 π] is obtained
[2π/3,2π]
According to the above conclusion
When x = 2 π / 3, the minimum value of Y is - 2
When x = 0, y = - 1, when x = 2 π, y = 1
∴ymax=1
The range of the function is [- 2,1]

Derivative problem of parametric equation function
From the parametric equation, x = aretant, y = in (the square of 1 + T)
Dy of DX = DX divided by dy of DX divided by DT = 1 divided by 2T of 1 + T squared = 2T
How does dy change into 2T divided by 1 + T squared? Isn't the Y derivative here 1 + T squared? How does 2T come out?

The derivation rule of compound function is used here
f'(g(x))=df/dg*g'(x)
In this problem, y = ln (1 + T ^ 2), after the derivation of LN, we have to further derive (1 + T ^ 2), so we get 2T

Y = 2Sin (2x + π / 3) (- π / 6 ≤ x ≤ π / 6)

Because - π / 6 ≤ x ≤ π / 6, 0 ≤ 2x + π / 3 ≤ 2 π / 3
So 0 ≤ 2Sin (2x + π / 3) ≤ 2

What is the derivative of y = Sin & # 178; xcos2x?

∵y＝（sinx）^2cos2x,
∴y′
＝［（sinx）^2］′cos2x＋（sinx）^2（cos2x）′
＝2sinxcosxcos2x－2sin2x（sinx）^2.

Y = 2Sin (π / 6-2x) (π / 3 ≤ x ≤ π) range

First deformation: F (x) = - 2Sin (2x - π / 6)
So when: 2K π + π / 2 ≤ 2x - π / 6 ≤ 2K π + 3 π / 2, that is: K π + π / 3 ≤ x ≤ K π + 5 π / 6,
F (x) is an increasing function;
So the increasing interval of F (x) is: [K π + π / 3, K π + 5 π / 6], K ∈ Z
So the range is (0,2]

What is the derivative of Sin & # x

Derivative = 2sinxcosx = sin2x

Y = √ 2cos ^ 2x + 5sinx-1 range

y=√2cos²x+5sinx-1
y²=2(1-sin²x)+5sinx-1
=-2sin²x+5sinx+1
=-2(sin²x-5/2sinx+25/16-25/16)+1
=-(sinx-5/4)²+33/8
∵y²≥0
So 0 ≤ - (sinx-5 / 4) & sup2; + 33 / 8 ≤ 33 / 8
0≤y≤√66 /4
So the range [0, √ 66 / 4]

Finding the derivative of sin (x / 2)

(sin(x/2))'=(1/2)cos(x/2).