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