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Question
\[\frac{3^x}{x + \tan x}\]
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Solution
\[\text{ Let } u = 3^x ; v = x + \tan x\]
\[\text{ Then }, u' = 3^x \log 3; v' = 1 + \sec^2 x\]
\[\text{ By quotient rule, we have }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{3^x}{x + \tan x} \right) = \frac{\left( x + \tan x \right) 3^x \log 3 - 3^x \left( 1 + \sec^2 x \right)}{\left( x + \tan x \right)^2}\]
\[ = \frac{3^x \left[ \left( x + \tan x \right) \log 3 - \left( 1 + \sec^2 x \right) \right]}{\left( x + \tan x \right)^2}\]
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