Let #y=f(x)=sqrt(x/(x+1)),# so, #x=4 rArr y=f(4)=sqrt(4/5)=2/sqrt5.#

So, we reqiure the eqn. of normal to the curve : #y=f(x)=sqrt(x/(x+1)),# at the pt. #(4,2/sqrt5)#

We recall that #[dy/dx]_(x=4,y=2/sqrt5)=f'(4)# gives us the slope tgt. line to the given curve.

Now, to diff. y, we can use the Quotient Rule. Instead, have a look at these two methods :-

**METHOD I :-**

Taking log. of both sides of the given eqn, we get, #lny=1/2{lnx-ln(x+1)} rArr d/dx(lny)=1/2d/dx{lnx-ln(x+1)} rArr d/dy(lny)*dy/dx=1/2{1/x-1/(x+1)}=1/{2x(x+1)} rArr 1/y*dy/dx=1/{2x(x+1)} rArr dy/dx=y/{2x(x+1)}#

#:. [dy/dx]_(x=4,y=2/sqrt5)=f'(4)=(2/sqrt5)/(2*4*5)=1/(20sqrt5).#

**METHOD II :-**

We write the eqn. of the given curve : #y^2(x+1)=x rArr d/dxy^2(x+1)=d/dxx rArr y^2*d/dx(x+1)+(x+1)*d/dxy^2=1 rArr y^2+(x+1){d/dyy^2}dy/dx=1 rArr y^2+2y(x+1)dy/dx=1 rArr x/(x+1)+2y(x+1)dy/dx=1......, [as, y^2=x/(x+1)] rArr 2y(x+1)dy/dx=1-x/(x+1)=1/(x+1) rArr dy/dx=1/{2y(x+1)^2}#

#[dy/dx]_(x=4,y=2/sqrt5)=f'(4)=1/{2*(2/sqrt5)*25}=1/(20sqrt5),# as before!

Since normal line is perp. to tgt., its slope is #-1/(f'(4))=-20sqrt5,.# and it passes thro. pt. #(4,2/sqrt5),# its eqn. is #: y-2/sqrt5=-20sqrt5(x-4)# or, #sqrt5y-2+100x-400=0,# i.e., #100x+sqrt5y-402=0.#