If $y = \cos 2x\,\,\cos 3x$, then ${y_n}$ is equal to

${6^n}\cos \left( {2x + \frac{{n{\rm{ }}\pi }}{2}} \right)\cos \left( {3x + \frac{{n{\rm{ }}\pi }}{2}} \right)$

${6^n}\cos \left( {2x + \frac{{n{\rm{ }}\pi }}{2}} \right)\cos \left( {\frac{{3x + n{\rm{ }}\pi }}{2}} \right)$

$\frac{1}{2}\left\{ {{5^n}\sin \left( {5x + \frac{{n{\rm{ }}\pi }}{2}} \right) + \sin \left( {x + \frac{\pi }{2}} \right)} \right\}$

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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

y = \cos 2x\,\,\cos 3x

, then

{y_n}

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the nth derivative of the product of cosine functions involves a sum of cosine functions with arguments adjusted by nπ/2, and the given options do not match the correct form.

JEE MATHEMATICS MCQ: DIFFERENTIATION

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