Finding the derivative of y = cos2x

Finding the derivative of y = cos2x

Here y = COSU, u = 2x
So y '= - sinu * u' = - sin (2x) * 2 = - 2sin2x

What is the derivative of y = cos2x

Let 2x = t
y=cost
y'=cost'*t'=-2sin2x

The derivative of y = sin ^ 3 * 2x?

y=[sin(2x)]^3
Let u = sin2x
So y = u ^ 3
So y '= 3U ^ 2 * u'
U '= 2cos2x
So y '= 6 (sin2x) ^ 2 * cos2x

How to find the derivatives of y = sin ^ 2 (2x + π / 3) and y = (sin (2x + π / 3)) ^ 2
Is the square of SiNx sin ^ 2 (x) or (sin (x)) ^ 2?

These two ways of writing are not the same... They are two identical functions
y`=sin^2(2x+π/3)`=2sin(2x+π/3)*((sin(2x+π/3)`)=2sin(2x+π/3)*cos(2x+π/3)*2=2sin(4x+2π/3)

There are simple steps to find the derivative of y = 2x * sin (2x + 5), but the more detailed the steps, the better. Thank you

Derivative of y = (sin 2x) / X

y'=[(sin2x)'*x-sin2x*x']/x²
=(2xcos2x-sin2x)/x²

Y = the derivative of e ^ sin (x ^ 2-1),

Find limit limx → 0 (x-sin2x / x + sin5x)

Original form
=lim[(1-sin2x/x)/(1+sin5x/x)]
=lim[(1-sin2x/(2x)*2)/(1+sin5x/(5x)*5)]
=[(1-lim(sin2x/(2x))*2)/(1+lim(sin5x/(5x))*5)]
=(1-2)/(1+5)
=-1/6

The limit of limx → 0 2sinx-sin2x / x ^ 3

lim(x→0) (2sinx-sin2x)/x^3
=lim(x→0) (2sinx-2sinxcosx)/x^3
=lim(x→0) 2sinx(1-cosx)/x^3
=lim(x→0) 2x*x^2/2*1/x^3
=1

Is limx → 0 | x | / X limited?

There is no limit, the left limit is - 1, the right limit is 1, the left and right limits are not equal, there is no limit