Correct Answer: Option D

Explanation

let y = \(\frac{x}{cosx}\) = x sec x ( since \(\frac{1}{cosx}\) = sec x )

let u =  x,   v = sec x

\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)

\(\frac{dy}{dx}\) =  x [secx tanx] + secx

Therefore,  \(\frac{dy}{dx}\)\(\frac{x}{cosx}\) = x secx tanx + secx

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