## Calculus 8th Edition

Equation of tangent: $3x+100y-52=0$ Equation of normal: $100x-3y-398.8=0$
$y=\frac{\sqrt x}{x+1}$. 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, $$y-0.4=\frac{100}{3}(x-4)\implies 3y-1.2=100x-400$$ $$100x-3y-398.8=0.$$