If $y=y(x)$ is an implicit function of $x$ such that $\log _{\mathrm{c}}(\mathrm{x}+\mathrm{y})=4 \mathrm{xy}$, then $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\mathrm{x}=0$ is equal to
$\ln (x+y)=4 x y \quad($ At $x=0, y=1)$
$x+y=e^{4 x y}$
$\Rightarrow 1+\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{4 \mathrm{xy}}\left(4 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}\right)$
At $x=0$
$\frac{d y}{d x}=3$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{e}^{4 \mathrm{xy}}\left(4 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}\right)^{2}+\mathrm{e}^{4 \mathrm{xy}}\left(4 \mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+4 \mathrm{y}\right)$
At $x=0, \frac{d^{2} y}{d x^{2}}=e^{0}(4)^{2}+e^{0}(24)$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=40$
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