If y = {≤ft( { x} )^{{{( x)}^{{{( x)}^.}{{^.}^.}^∞ }}}} then {{dy}}{{dx}} is equal to

Correct Answer is : −y2tan⁡x1−y In (cos⁡x)- {{{y² x}}{{1 - y\;In\;( x)}}−1−yIn(cosx)y2tanx

- {{{y² x}}{{1 - y\;In\;( x)}}

Correct Answer is : −y2tan⁡x1−y In (cos⁡x)- {{{y² x}}{{1 - y\;In\;( x)}}−1−yIn(cosx)y2tanx

Test Index

UPSC NDA 2 2017 Mathematics Solved Paper

Question : 63 of 120

Marks: +1, -0

y = {\left( {\cos x} \right)^{{{(\cos x)}^{{{(\cos x)}^.}{{^.}^.}^\infty }}}}

\frac{{dy}}{{dx}}

is equal to

Solution:

Concept:

For an infinite power tower, set

y = (\cos x)^y

and use logarithmic differentiation.

Explanation:

Given

y = (\cos x)^{(\cos x)^{.^{.^{.}}}}

We write

y = (\cos x)^y

Take natural logarithm:

\ln y = y \ln(\cos x)

Differentiate both sides with respect to

x

\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{-\sin x}{\cos x} + \frac{dy}{dx} \ln(\cos x)

Simplify:

\frac{1}{y} \frac{dy}{dx} = -y \tan x + \frac{dy}{dx} \ln(\cos x)

Group

\frac{dy}{dx}

terms:

\frac{dy}{dx}\left(\frac{1}{y} - \ln(\cos x)\right) = -y \tan x

Multiply both sides by

y

\frac{dy}{dx}\left(1 - y \ln(\cos x)\right) = -y^2 \tan x

Thus,

\frac{dy}{dx} = -\frac{y^2 \tan x}{1 - y \ln(\cos x)}

Answer:

Option A.

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