If $y=e^{4 x} \cos 5 x$, then $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is

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Aicharya · Accepted Answer

Given $y=e^{4 x} \cos 5 x$ $\begin{aligned} 	herefore & \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x})+\cos 5 \mathrm{x}\left(4 \mathrm{e}^{4 \mathrm{x}}ight) \quad=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \ & \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[\mathrm{e}^{4 \mathrm{x}}(-25 \cos 5 \mathrm{x}-20 \sin 5 \mathrm{x})ight]+\left[(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \cdot\left(4 \mathrm{e}^{4} \mathrm{x}ight)ight] \ &\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}ight)_{\mathrm{x}=0}=-25+(4 	imes 4)=-25+16=-9 \end{aligned}$

If $y=e^{4 x} \cos 5 x$, then $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is

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