Answer
Equation of tangent: $3x+100y-52=0$
Equation of normal: $100x-3y-398.8=0$
Work Step by Step
$y=\frac{\sqrt x}{x+1}$. Differentiating both sides with respect to $x$, at point $(4, 0.4)$, using quotient's rule,
$$\frac{dy}{dx}=\frac{(1/2)x^{-1/2}(x+1)-x^{1/2}}{(x+1)^2}$$
$$\frac{dy}{dx}\bigg|_{x=4}=\frac{(1/2)4^{-1/2}(4+1)-4^{1/2}}{(4+1)^2}=\frac{\frac{5}{4}-2}{25}=-\frac{3}{100}.$$
Thus, the slope of the tangent line is $-\frac{3}{100}$. Using point-slope form, the equation of the required tangent is,
$$y-0.4=-\frac{3}{100}(x-4)\implies 100y-40=-3x+12$$
$$3x+100y-52=0$$
Since, the slope of tangent, $m_1=-0.03$, then the slope of normal, $m_2$, can be found out using the relation, $m_1\times m_2=-1$. Thus, $m_2=100/3$. Using point-slope form, the equation of required normal is,
$$y-0.4=\frac{100}{3}(x-4)\implies 3y-1.2=100x-400$$
$$100x-3y-398.8=0.$$
