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 $ $ 2\text{ }cos\text{ }x\text{ }cos\text{ }3x=cos\text{ }4x+cos\text{ }2x $ $ \therefore $ $ \frac{d}{dx}(~2\text{ }cos\text{ }x\text{ }cos\text{ }3x)=\frac{d}{dx}(cos\text{ }4x+cos\text{ }2x) $ $ =-4\text{ }sin\text{ }4x-2\text{ }sin\text{ }2x $ $ \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) $