What is the equation of the line normal to # f(x)=sqrt(lnx)# at # x=4#?

1 Answer
Ratnaker Mehta
Jun 24, 2016

#(8sqrtln4)x+y-33sqrtln4=0.#

Explanation:

#y=f(x)=sqrtlnx=(lnx)^(1/2)#
#:. f'(x)=1/2(lnx)^(1/2-1)(lnx)'#.............[Chain Rule]
#=1/(2sqrtlnx)(1/x)=1/(2xsqrtlnx).#

#f'(4)=1/(8sqrtln4).#

Recall that#f'(4)# is the slope of tgt. line to #f# at #x=4.#

Normal is perp. to tgt., hence, slope of normal #=-1/(f'(4))=-8sqrtln4.#

It passes thro. pt. #(4,f(4))=(4,sqrtln4)#

#:.# The eqn. of normal is, #y-sqrtln4=-(8sqrtln4)(x-4),# written in std. form, #(8sqrtln4)x+y-32sqrtln4-sqrtln4=0,.# i.e., #(8sqrtln4)x+y-33sqrtln4=0.#

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