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Harshit Sahu Grade: 11
        

cosx.cos2x.cos4x.cos8x.cos16x

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

Y = cosxcos2xcos4xcos8xcos16x


taking log both sides


 


logY = log(cosx.cos2x.cos4x.cos8x.cos16x)




logab = loga + logb        (property)




logY =[ logcosx + logcos2x + logcos4x + logcos8x + logcos16x ]           ...................1


differentiating wrt x     


differentiatn of log(f(x) = f1x/ f(x)




(1/Y) . (y1)  =  


                          [ -sinx/cosx + -sin2x/2cos2x - sin4x/4cos4x - sin8x/8cos8x                                                 -16sin16x/16cos16x]


 y1 = -Y[tanx + tan2x/2 + tan4x/4 + tan8x/8 + tan16x/16] .....................2


 


now put the value of Y from 1 in eq 2 u will get the desired result...



6 years ago
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