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Differentiate the following w.r.t. x : at[(tanx)tanx]tanxat x=π4

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प्रश्न

Differentiate the following w.r.t. x : `[(tanx)^(tanx)]^(tanx) "at"  x = pi/(4)`

योग
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उत्तर

Let y = `[(tanx)^(tanx)]^(tanx)`
∴ log y = `log[(tanx)^(tanx)]tanx`
= tanx. log(tanx)tanx
= tanx. tanx log(tan x)
= (tanx)2. log(tan x)
Differentiating both sides w.r.t. x, we get
`1/y."dy"/"dx" = "d"/"dx"[tanx)^2.log(tanx)]`

= `(tanx)^2."d"/"dx"(log tanx) + (log tanx)."d"/"dx"(tanx)^2`

= `(tanx)^2. xx 1/tanx."d"/"dx"(tanx) + (log tanx) xx 2tanx."d"/"dx"(tanx)`

= `(tanx)^2 xx 1/tanx.sec^2x + (log tanx) xx 2 tanxsec^2x`

∴ `"dy"/"dx" = y[(tanx)(sec^2x) + (logtanx)(2tanxsec^2x)]`

= [(tanx)tanx]tanx.(tanxsec2x)[1 + 2logtanx]

If x = `pi/(4)`, then

`"dy"/"dx" = [(tan pi/4)^(tan  pi/4)]^(tan  pi/4)(tan  pi/4 sec^2  pi/4)[1 + 2log tan  pi/4]`
= `[(1)^1]^1.[1(sqrt(2))^2][1 + 2log1]`
= 1 x 2 x 1                  ...[∵ log 1 = 0]
= 2.

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अध्याय 1: Differentiation - Exercise 1.3 [पृष्ठ ४०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.3 | Q 2.8 | पृष्ठ ४०

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