first principal

Note that :

 sin^2(2x)   = [1 - cos(4x)]/ 2     ......so we have.....

[ 1 - cos4(x + h)] / 2] * [ 1/h]    -    [1 - cos(4x)] / 2 * [ 1/h ] =

[1 - cos(4x + cos4h)] / 2h   - [1 - cos(4x)] 2h  =

[cos(4x)  -  cos(4x + 4h) ] / 2h   =

[ cos(4x)  - [cos4xcos4h  - sin4xsin4h] ] / 2h  =

[ cos4x ][1 - cos4h] / 2h    +  sin4xsin4h/ 2h        [multiply top/bottom of both fractions by 2]

[2cos4x] [(1 - cos4h) / 4h ] +   [2sin4x] [sin4h/ 4h ]       let h →  0

[2cosx * 0]  +  [2sin4x * 1 ]  =

2sin4x  =

2sin(2x + 2x)  =

2[ sin2xcos2x + sin2xcos2x]  =

2 [ 2 * sin2xcos2x]  =

4[sin2x] [cos2x]

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Note that :   [sin(2x)]^2  =  [sin2x ] * [sin2x]     (1)

And using the Product Rule, the derivative  of (1)  =

[2cos2x] [sin2x] + [sin2x][ 2 cos2x]  =

[2cos2x] [ sin2x + sin2x]  =

[2cos2x] [2sin2x] =

4[sin2x][cos2x]

Find the derivative of the following via implicit differentiation:

d/dx(f(x)) = d/dx(sin^2(2 x))

The derivative of f(x) is f'(x):

f'(x) = d/dx(sin^2(2 x))

Using the chain rule, d/dx(sin^2(2 x)) = ( du^2)/( du) ( du)/( dx), where u = sin(2 x) and ( d)/( du)(u^2) = 2 u:

f'(x) = 2 d/dx(sin(2 x)) sin(2 x)

Using the chain rule, d/dx(sin(2 x)) = ( dsin(u))/( du) ( du)/( dx), where u = 2 x and ( d)/( du)(sin(u)) = cos(u):

f'(x) = cos(2 x) d/dx(2 x) 2 sin(2 x)

Factor out constants:

f'(x) = 2 d/dx(x) 2 cos(2 x) sin(2 x)

Simplify the expression:

f'(x) = 4 cos(2 x) (d/dx(x)) sin(2 x)

The derivative of x is 1:

f'(x) = 1 4 cos(2 x) sin(2 x)

Simplify the expression:

f'(x) = 4 cos(2 x) sin(2 x)

Expand the left hand side:

Answer: |f'(x) = 4 cos(2x) sin(2x) =f'(x) = 2 sin(4x)