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Thanks in dac

Veteran  May 25, 2017
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 #1
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Possible derivation:
d/dx(tan^(-1)(sin(5 x)))
Using the chain rule, d/dx(tan^(-1)(sin(5 x))) = ( dtan^(-1)(u))/( du) ( du)/( dx), where u = sin(5 x) and ( d)/( du)(tan^(-1)(u)) = 1/(1 + u^2):
 = (d/dx(sin(5 x)))/(1 + sin^2(5 x))
Using the chain rule, d/dx(sin(5 x)) = ( dsin(u))/( du) ( du)/( dx), where u = 5 x and ( d)/( du)(sin(u)) = cos(u):
 = cos(5 x) d/dx(5 x)/(1 + sin(5 x)^2)
Factor out constants:
 = (5 d/dx(x) cos(5 x))/(1 + sin^2(5 x))
The derivative of x is 1:
Answer: | = (15 cos(5x)) / (1 + sin^2(5x))

Guest May 25, 2017
 #2
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+1

 

 

Note  .......derivative of arctan (u)  =   1 / ( 1 + u^2) * du  .....so......

 

f(x)  =   arctan (sin (5x))   =   [  1 /  [ 1 + (sin(5x) )^2]  * 5 cos (5x)  =

 

[ 5 cos (5x) ]  /  [ 1 + sin2(5x) ]

 

 

cool cool cool

CPhill  May 25, 2017

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