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