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Question

If y=[(tanx)tanx]tanx, then dydx at x=π/4 is

A
2
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B
1/2
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C
1
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D
0
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Solution

The correct option is D 2

Consider the following differential eq.

y=((tanx)tanx)tanx&x=π4

Taking log both side, we get.

logy=tan2xlog(tanx)

Differentiate both side w.r.t x, we get.

1ydydx=2tanxsec2xlog(tanx)+tan2x1tanxsec2x

1ydydx=((tanx)tanx)tanx(2tanπ4sec2π4log(tanπ4)+tanπ4sec2π4)

dydx=0+2

=2

Hence, this is the required answer.


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