Question

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

Answer 1

`(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.`