Question

If $y=\cos 2x\cos 3x$, then $y_n$ is equal to
Where, $y_n$ denotes the $n^{th}$ derivative of $y$.

• A. $6^n\cos (2x+\frac{n\pi }{2})\cos (3x+\frac{n\pi }{2})$
• B. $\frac{1}{2}[5^n\cos (\frac{n\pi }{2}+5x)+\cos (\frac{n\pi }{2}+x)]$
• C. $\frac{1}{2}[5^n\sin (5x+\frac{n\pi }{2})+\sin (x+\frac{\pi }{2})]$
• D. $0$

Answer 1

Answer: (B)

$\frac{1}{2}[5^n\cos (\frac{n\pi }{2}+5x)+\cos (\frac{n\pi }{2}+x)]$

$y=\frac{1}{2}\left(2\cos 2x\cos 3x\right)$
$y=\frac{1}{2}(\cos 5x+\cos x)$
$y_1=\frac{1}{2}(-5\sin 5x-\sin x)$
$y_1=\frac{1}{2}(5\cos (\frac{\pi }{2}+5x)+\cos (\frac{\pi }{2}+x))$
$y_2=\frac{1}{2}(-5^2\sin (\frac{\pi }{2}+5x)-\sin (\frac{\pi }{2}+x))$
$y_2=\frac{1}{2}\left(5^2\cos \right(\pi +5x)+\cos (\pi +x\left)\right)$
$y_3=\frac{1}{2}(5^3\cos (\frac{3\pi }{2}+5x)+\cos (\frac{3\pi }{2}+x))$
It is observed that with each consecutive derivative, $\frac{\pi }{2}$ can be added to the argument of both terms in the bracket for generalization.
$\therefore y_n=\frac{1}{2}(5^n\cos (\frac{n\pi }{2}+5x)+\cos (\frac{n\pi }{2}+x))$