We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
986
2
avatar

use first principal to find the derivative of 

f(x)=sin^2(2x)

\(f(x)=sin^2(2x)\)

 Feb 13, 2016

Best Answer 

 #2
avatar+104712 
+5

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]

 

-------------------------------------------------------------------------

 

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]

 

 

cool cool cool

 Feb 13, 2016
edited by CPhill  Feb 13, 2016
edited by CPhill  Feb 13, 2016
edited by CPhill  Feb 13, 2016
 #1
avatar
+5

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)

 Feb 13, 2016
 #2
avatar+104712 
+5
Best Answer

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]

 

-------------------------------------------------------------------------

 

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]

 

 

cool cool cool

CPhill Feb 13, 2016
edited by CPhill  Feb 13, 2016
edited by CPhill  Feb 13, 2016
edited by CPhill  Feb 13, 2016

28 Online Users

avatar
avatar