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Question

If y=cosnx.sinnxand dydx=n.cosn-1x×cosB then B is equals to ?


A

(n1)x

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B

(n+1)x

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C

nx

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D

(1n)x

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Solution

The correct option is B

(n+1)x


Explanation for the correct option:

Find the value of B:

Given functions,

y=cosnx.sinnx

Differentiate the function y=cosnx·sinnx with respect to x:

dydx=n·cosn1x·(sinx)·sinnx+ncosnx·cosnx(n)[d(cosx)dx=-sinx,d(sinx)dx=cosx]=ncosn1x·sinnx·(sinx)+ncosnxcosnx=ncosnx(sinx·sinnx)cosx+cosnx=ncosnx(sinxsinnx+cosxcosnx)cosx=n·cosn1x·cos(x+nx)

Compare the obtained result with dydx=n.cosn-1x×cosB

B=x+nx=x(n+1)

Hence, the option (B) is the correct.


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