Question

$\frac{d^{20}}{d{x}^{20}}(2\cos x\cos 3x)$ is equal to:

• A. $2^{20}(\cos 2x+2^{20}\cos 4x)$
• B. $2^{20}(\cos 2x-2^{20}\sin 2x)$
• C. $-6\sin x\sin 3x$
• D. $6\sin x\sin 3x$
• E. $6(2^{20}\cos 2x-2^{40}\cos 4x)$

Answer 1

Answer: (A)

$\therefore$ $2cosxcos3x=cos4x+cos2x$ $\therefore$ $\frac{d}{dx}(2cosxcos3x)=\frac{d}{dx}(cos4x+cos2x)$ $=-4sin4x-2sin2x$ $\frac{d^2}{d{x}^2}(2\cos x\cos 3x)=-16\cos 4x-4cos2x$ $\therefore$ $\frac{d^{20}}{d{x}^{20}}(2\cos x\cos 3x)$ $=4^{20}\cos 4x+{(2)}^{20}\cos 2x$ $=2^{20}(2^{20}\cos 4x+\cos 2x)$