ln(x + y) = 4xy (At x = 0, y = 1) x + y = e 4xy $\Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$ At x = 0 ${{dy} \over {dx}} = 3$ ${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$ At x = 0, ${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$ $\Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$