d^2 d x^2 (2 x , 3x) = - Vidyadip

MCQ

${{{d^2}} \over {d{x^2}}}(2\cos x\,\cos 3x) = $

A
${2^2}(\cos 2x + {2^2}\cos 4x)$
B
${2^2}(\cos 2x - {2^2}\cos 4x)$
C
${2^2}( - \cos 2x + {2^2}\cos 4x)$
✓
$ - {2^2}(\cos 2x + {2^2}\cos 4x)$

✓

Answer

Correct option: D.

$ - {2^2}(\cos 2x + {2^2}\cos 4x)$

d
(d) $y = 2\cos x\cos 3x$

$\frac{{dy}}{{dx}} = 2\cos x\,.\,( - 3\sin 3x) + 2\cos 3x( - \sin x)$

$ = \, - 3(\sin 4x + \sin 2x) + ( - 1)[\sin 4x + \sin ( - 2x)]$

$\frac{{{d^2}y}}{{d{x^2}}} = - 3(4\cos 4x + 2\cos 2x) - 1(4\cos 4x - 2\cos 2x)$

$ = - 16\cos 4x - 4\cos 2x$$ = - 4(\cos 2x + 4\cos 4x)$

$ = - {2^2}(\cos 2x + {2^2}\cos 4x)$.

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