If y = y ( x ) is an implicit function of x such that _ e ( x + y )...

Given:

$\log_{e} ( x + y ) = 4xy$

When $x = 0$, then $y = 1$

$\log_{e}\left(x + y\right) = 4xy$

$⇒ x + y = {e}^{4xy}$

Now differentiate w.r.t. $x$

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

At $( 0, 1 ) ⇒ {y}^{'} ( 0 ) + 1 = 4 ⇒ {y}^{'} ( 0 ) = 3$

Now, again differentiate equation $( i )$, we get

${y}^{"} = {e}^{4xy}{\left(4y + 4x{y}^{′}\right)}$^{2} + {e}^{4xy}\left(4{y}^{'} + 4{y}^{'} + 4x{y}^{"}\right)$</p><p>At \left(0, 1\right)$

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

$⇒ {y}^{"} ( 0 ) = 16 + 24 = 40$

$⇒ {y}^{"} ( 0 ) = 40$