Find the equation of a normal to the curve

finding the slope of the tangent by differentiating the curve

$\frac{\mathrm{dy}}{\mathrm{dx}}=\ln \mathrm{x}+1$

$m$ (tangent) $=\ln x+1$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$]

$m($ normal $)=-\frac{1}{\ln x+1}$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

now comparing the slope of normal with the given equation

$\mathrm{m}($ normal $)=1$

$-\frac{1}{\ln x+1}=1$

$x=\frac{1}{e^{2}}$

since this point lies on the curve, we can find $y$ by substituting $x$

$y=-\frac{2}{e^{2}}$

The equation of normal is given by

$y+\frac{2}{e^{2}}=x-\frac{1}{e^{2}}$