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प्रश्न
\[\frac{d^{20}}{d x^{20}} \left( 2 \cos x \cos 3 x \right) =\]
विकल्प
220 (cos 2 x − 220 cos 4 x)
220 (cos 2 x + 220 cos 4 x)
220 (sin 2 x + 220 sin 4 x)
220 (sin 2 x − 220 sin 4 x)
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उत्तर
(b) 220(cos2x + 220cos4x)
Here,
\[y = 2\cos x \cos3x = \cos\left( 3x - x \right) + \cos\left( 3x + x \right)\]
\[ = \cos2x + \cos4x\]
\[ \Rightarrow \frac{d y}{d x} = - 2 \sin2x - 4 \sin4x = - 2\left( \sin2x + 2 \sin4x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - 4 \cos2x - 16 \cos4x = - 2^2 \left( \cos2x + 2^2 \cos4x \right)\]
\[ \Rightarrow \frac{d^3 y}{d x^3} = 2^3 \left( \sin2x + 2^3 \sin4x \right)\]
\[ \Rightarrow \frac{d^4 y}{d x^4} = 2^3 \left( 2\cos2x + 4 \times 2^3 \cos4x \right) = 2^4 \left( \cos2x + 2^4 \cos4x \right)\]
\[ \therefore \frac{d^{20} \left( \cos2x + \cos4x \right)}{d x^{20}} = 2^{20} \left( \cos2x + 2^{20} \cos4x \right)\]
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