If y=cos⁡6x+6cos⁡4x+15cos⁡2x+10cos⁡5x+5cos⁡3x+10cos⁡x, then dydx is

y=cos⁡6x+6cos⁡4x+15cos⁡2x+10cos⁡5x+5cos⁡3x+10cos⁡xy=cos⁡6x+cos⁡4x+5cos⁡4x+5cos⁡2x+10(1+cos⁡2x)cos⁡5x+5cos⁡3x+10cos⁡x⇒y=2cos⁡x⋅cos⁡5x+5⋅2⋅cos⁡3x⋅cos⁡x+2⋅10cos2⁡xcos⁡5x+5cos⁡3x+10cos⁡xy=2cos⁡x(cos⁡5x+5cos⁡3x+10cos⁡x)cos⁡5x+5cos⁡3x+10cos⁡x⇒y=2cos⁡x ⇒ dydx=−2sin⁡x

If y=cos⁡6x+6cos⁡4x+15cos⁡2x+10cos⁡5x+5cos⁡3x+10cos⁡x, then dydx is

y=cos⁡6x+6cos⁡4x+15cos⁡2x+10cos⁡5x+5cos⁡3x+10cos⁡xy=cos⁡6x+cos⁡4x+5cos⁡4x+5cos⁡2x+10(1+cos⁡2x)cos⁡5x+5cos⁡3x+10cos⁡x⇒y=2cos⁡x⋅cos⁡5x+5⋅2⋅cos⁡3x⋅cos⁡x+2⋅10cos2⁡xcos⁡5x+5cos⁡3x+10cos⁡xy=2cos⁡x(cos⁡5x+5cos⁡3x+10cos⁡x)cos⁡5x+5cos⁡3x+10cos⁡x⇒y=2cos⁡x ⇒ dydx=−2sin⁡x

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If $y = \frac{\cos 6 x + 6 \cos 4 x + 15 \cos 2 x + 10}{\cos 5 x + 5 \cos 3 x + 10 \cos x},$ then $\frac{d y}{d x}$ is

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$2 \sin x + \cos x$

$- 2 \sin x$

$\cos 2 x$

$\sin 2 x$

answer is B.

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Detailed Solution

$y = \frac{\cos 6 x + 6 \cos 4 x + 15 \cos 2 x + 10}{\cos 5 x + 5 \cos 3 x + 10 \cos x}$

$\begin{aligned} y = \frac{\cos 6 x + \cos 4 x + 5 \cos 4 x + 5 \cos 2 x + 10 (1 + \cos 2 x)}{\cos 5 x + 5 \cos 3 x + 10 \cos x} \\ \Rightarrow y = \frac{2 \cos x \cdot \cos 5 x + 5 \cdot 2 \cdot \cos 3 x \cdot \cos x + 2 \cdot 10 \cos^{2} x}{\cos 5 x + 5 \cos 3 x + 10 \cos x} \end{aligned}$

$\begin{aligned} y = \frac{2 \cos x (\cos 5 x + 5 \cos 3 x + 10 \cos x)}{\cos 5 x + 5 \cos 3 x + 10 \cos x} \\ \Rightarrow y = 2 \cos x \Rightarrow \frac{d y}{d x} = - 2 \sin x \end{aligned}$