If y = y ( x ) is an implicit function of x such that log e ( x + y ) = 4 x y , then d 2 y   d x 2 at x…

Given:

${\log }_e(x+y)=4xy$

When $x=0$, then $y=1$

${\log }_e\left(x+y\right)=4xy$

$\Rightarrow x+y=e^{4xy}$

Now differentiate w.r.t. $x$

$1+y^{\text{'}}=e^{4xy}\left(4y+4x{y}^{\text{'}}\right) \dots \left(\mathrm{i}\right)$

At $(0,1)\Rightarrow y^{\text{'}}\left(0\right)+1=4\Rightarrow y^{\text{'}}\left(0\right)=3$

Now, again differentiate equation $\left(\mathrm{i}\right)$, we get

$y^{"}=e^{4xy}{\left(4y+4x{y}^{\mathrm{\prime }}\right)}^2+e^{4xy}\left(4y^{\text{'}}+4y^{\text{'}}+4x{y}^{"}\right)$

At $\left(0,1\right)$

$y^{"}\left(0\right)=1(4\times 1+0{)}^2+1(4\times 3+4\times 3+0)$

$\Rightarrow y^{"}\left(0\right)=16+24=40$

$\Rightarrow y^{"}\left(0\right)=40$