Write as #y=(3x+2)^(1/2)#

Using calculus

Let #u=3x+2 => (du)/(dx) = 3#....(1)

By substitution

#y=u^(1/2) => (dy)/(du)=1/2u^(-1/2)# ....(2)

By comparing equations (1) and (2)

#(dy)/(dx)=(dy)/(cancel(du))xxcancel(du)/(dx)#

#=>(dy)/(dx)=1/2u^(-1/2)xx3#

#=>(dy)/(dx)=3/(2sqrt(3x+2))#

Multiply by 1 but in the form of #1=sqrt(3x+2)/sqrt(3x+2)#

#dy/(dx) =(3sqrt(3x+2))/(3x+2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine gradient of normal")#

Condition for #x=1#

Gradient of given curve #-> (3sqrt(5))/5#

#"So gradient of normal is "-> -5/(3sqrt(5)) #

Multiply by 1 but in the form of #1=(sqrt(5))/(sqrt(5))# giving:

#" "-(cancel(5)sqrt(5))/(3xxcancel(5))#

#color(brown)("gradient "->-(sqrt(5))/3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine point through which the normal passes")#

For #x=1#

#y=sqrt(3x+2) " "vec("becomes")" " y=sqrt(5)#

#=> "Point "->(x,y)->(1,sqrt(5))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine equation of normal")#

#y=-1/mx+c" "vec("becomes")" " sqrt(5)=(-sqrt(5)/3xx1)+c#

add#" "sqrt(5)/3# to both sides

#sqrt(5)+sqrt(5)/3=0+c#

#=> c= (4sqrt(5))/3#

#color(blue)(y=-sqrt(5)/3 x + (4sqrt(5))/3color(white)(2/2)#