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If $y=\cos ^2\left(\frac{5 x}{2}\right)-\sin ^2\left(\frac{5 x}{2}\right)$, then $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=$
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The correct answer is:
-25y
$y=\cos ^2 \frac{5 x}{2}-\sin ^2 \frac{5 x}{2}=\cos 5 x$
$\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-5 \sin 5 x$
$\Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-25 \cos 5 x=-25 y$
$\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-5 \sin 5 x$
$\Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=-25 \cos 5 x=-25 y$
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