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KEAM 2021 Math Solved Paper

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Question : 56 of 120

Marks: +1, -0

Let

y=(\tan x)^{\sin x}

0 < x < \frac{\pi}{2} .

\frac{dy}{dx}=(\tan x)^{\sin x}((\cos x) \log (\tan x)+g(x))

g(x)=

$\sin x \sec ^{2} x$
$\sec x \csc x$
$\csc x$
$\sin x \tan x$
$\sec x$

Solution:

y=(\tan x)^{\sin x}

Taking log on both sides,

\log y=\log (\tan x)^{\sin x}

\log y=\sin x \log (\tan x)

Differentiating both side w.r.t.

x

,

\frac{1}{y} \frac{dy}{dx}=\cos x \cdot \log (\tan x)+\frac{\sin x}{\tan x} \times \sec ^{2} x

\frac{dy}{dx}=y[\cos x \cdot \log (\tan x)+\sec x]

\frac{dy}{dx}=(\tan x)^{\sin x}(\cos x \log (\tan x)+\sec x)

Comparing the value of

\frac{dy}{dx}

with

(\tan x)^{\sin x} \cdot[\cos x \log (\tan x)+g(x)]

Hence,

g(x)=\sec x

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