If f (x) has a minimum f (a) and a maximum f (b), then f (a) must be less than f (b)? The content of derivative

If f (x) has a minimum f (a) and a maximum f (b), then f (a) must be less than f (b)? The content of derivative

not always
If it is a constant function, the two are equal

Finding the derivative of y = cos (X-2)
There should be a detailed process. Thank you

Formula: (cosx) '= - SiNx
Y '= - sin (X-2) * (X-2)' (derivation of composite function)
=-sin(x-2)*1
=-sin(x-2)

How to calculate the derivative of y = cos (2 π - x)

y=cos(2π-x）
=cosx
y'=(cosx)'=-sinx

Finding the n-th derivative of y = cos ^ 2 (x)
The answer is 2 ^ (n-1) cos (2x + n π / 2),

The derivative of COS ^ 2 (x)
I know the answer is - sin2x
What I want to ask is:
For a function, as long as there is a cusp in the image, there is no derivative
Isn't there a cusp in the image of COS ^ 2 (x)?
(isn't the shape of the image of the function just to turn up all the parts below the x-axis of the cosx image? Aren't there all sharp points where x = k π + 0.5 π?)

A:
y=(cosx)^2
y'(x)=2cosx*(-sinx)=-sin2x
y=(cosx)^2=(1/2)(cos2x+1)=(1/2)cos2x+1/2
So: there is no cusp problem with this function
|Cosx | plus absolute value, just like the building owner said, there is a sharp roof problem,
Fold the image below the x-axis up and there is a tip

How to find the derivative of COS (x + y)

The partial derivative of X is obtained
-sin(x+y)
For y,
-sin(x+y)

Derivative of y = cos (x ^ 2 + X + 1)

-(2x+1)sin(x^2+x+1)

Finding the n-th derivative of y = x ^ 3 SiNx

y′=3x² sinx + x³cosxy〃=6xsinx + 3x²cosx +3x²cosx -x³sinx=6xsinx + 6x²cosx -x³sinxy(³)=6sinx +6xcosx+12xcosx-6x²sinx-3x²sinx-x³cosx=6sinx +18xco...

Finding the n-order derivative of y = SiNx

When n = 2K + 1, it is equal to (1) k Power cosx
When n = 2K, it is equal to (- 1) k Power SiNx

Finding the n-th derivative of y = (SiNx) ^? (with process)
The Nth derivative of y = (SiNx) ^ 2

Y = Sin & # 178; X = (1 / 2) (1-cos2x) y '= (1 / 2) * 2Sin (2x) = sin (2x) y' '= 2cos (2x) = 2Sin (2x + π / 2) y' '= - 4sin (2x) = 4sin (2X + π) y ^ (4) = - 8cos (2x) = 8sin (2x + 3 π / 2) y ^ (5) = 16sin (2x) = 16sin (2x + 2 π). Y ^ (n) = [2 ^ (n-1)] sin (2x + (n-1) π / 2) hope that