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Question

If y=(tanx)(tanx)(tanx), then find the value of dydx at x=π4

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Solution

y=tanxtanxtanx................ .......(i)


From (i), we have

y=tanxy
Taking log on both sides

logy=ylogtanx

Differentiating on both sides, we get

1ydydx=ysec2xtanx+logtanxdydx

Put x=π4 and y=1

11dydx=(1)sec2π4tanπ4+logtanπ4dydx

dydx=21+log1dydx

dydx=2+0dydx

dydx=2

So, correct answer is 2.


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