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If $y=e^{4 x} \cos 5 x$, then $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is
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Verified Answer
The correct answer is:
$-9$
Given $y=e^{4 x} \cos 5 x$
$\begin{aligned}
\therefore & \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x})+\cos 5 \mathrm{x}\left(4 \mathrm{e}^{4 \mathrm{x}}\right) \quad=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \\
& \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[\mathrm{e}^{4 \mathrm{x}}(-25 \cos 5 \mathrm{x}-20 \sin 5 \mathrm{x})\right]+\left[(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \cdot\left(4 \mathrm{e}^{4} \mathrm{x}\right)\right] \\
&\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=0}=-25+(4 \times 4)=-25+16=-9
\end{aligned}$
$\begin{aligned}
\therefore & \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x})+\cos 5 \mathrm{x}\left(4 \mathrm{e}^{4 \mathrm{x}}\right) \quad=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \\
& \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[\mathrm{e}^{4 \mathrm{x}}(-25 \cos 5 \mathrm{x}-20 \sin 5 \mathrm{x})\right]+\left[(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \cdot\left(4 \mathrm{e}^{4} \mathrm{x}\right)\right] \\
&\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=0}=-25+(4 \times 4)=-25+16=-9
\end{aligned}$
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