If y=a cos (sin 2 x)+b sin (sin 2 x), then y^ | vedclass

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(C) Given, $y=a \cos (\sin 2 x)+b \sin (\sin 2 x)$.On differentiating both sides with respect to $x$, we get:$y^{\prime} = -a \sin(\sin 2x) \cdot \cos

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$-4(\cos ^2 2 x) y$

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(C) Given, $y=a \cos (\sin 2 x)+b \sin (\sin 2 x)$.On differentiating both sides with respect to $x$, we get:$y^{\prime} = -a \sin(\sin 2x) \cdot \cos 2x \cdot 2 + b \cos(\sin 2x) \cdot \cos 2x \cdot 2$$y^{\prime} = 2 \cos 2x \{-a \sin(\sin 2x) + b \cos(\sin 2x)\}$.Again, differentiating with respect to $x$ using the product rule:$y^{\prime \prime} = -4 \sin 2x \{-a \sin(\sin 2x) + b \cos(\sin 2x)\} + 2 \cos 2x \{-a \cos(\sin 2x) \cdot \cos 2x \cdot 2 - b \sin(\sin 2x) \cdot \cos 2x \cdot 2\}$.Substituting $y^{\prime} = 2 \cos 2x \{-a \sin(\sin 2x) + b \cos(\sin 2x)\}$ into the first part:$y^{\prime \prime} = -4 \sin 2x \cdot \frac{y^{\prime}}{2 \cos 2x} - 4 \cos^2 2x \{a \cos(\sin 2x) + b \sin(\sin 2x)\}$.Since $y = a \cos(\sin 2x) + b \sin(\sin 2x)$, we have:$y^{\prime \prime} = -2 	an 2x \cdot y^{\prime} - 4 \cos^2 2x \cdot y$.Therefore, $y^{\prime \prime} + (2 	an 2x) y^{\prime} = -4 \cos^2 2x \cdot y$.

If $y=a \cos (\sin 2 x)+b \sin (\sin 2 x)$, then $y^{\prime \prime}+(2 \tan 2 x) y^{\prime}$ is equal to

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