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Mathematics · JEE | LIMITS AND DERIVATIVES

हिन्दी

Find the derivative of \( f(x) = \frac{x + \cos x}{\tan x} \).

  1. A
    \( \frac{(1 - \sin x)\tan x - (x + \cos x)\sec^2 x}{\tan^2 x} \)
  2. B
    \( \frac{(1 + \sin x)\tan x + (x + \cos x)\sec^2 x}{\tan^2 x} \)
  3. C
    \( \frac{\sin x \tan x - (x + \cos x)\sec^2 x}{\tan^2 x} \)
  4. D
    \( \frac{(1 - \sin x)\tan x + (x + \cos x)\sec^2 x}{\tan^2 x} \)
View Answer & Solution

Correct answer · Option A

\( \frac{(1 - \sin x)\tan x - (x + \cos x)\sec^2 x}{\tan^2 x} \)

Explanation

Applying the quotient rule to \( \frac{u}{v} \) with \( u = x + \cos x \) and \( v = \tan x \) gives \( \frac{u'v - uv'}{v^2} = \frac{(1 - \sin x)\tan x - (x + \cos x)\sec^2 x}{\tan^2 x} \).
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