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

Jun 24, 2016

$\left(8 \sqrt{\ln} 4\right) x + y - 33 \sqrt{\ln} 4 = 0.$

#### Explanation:

$y = f \left(x\right) = \sqrt{\ln} x = {\left(\ln x\right)}^{\frac{1}{2}}$
$\therefore f ' \left(x\right) = \frac{1}{2} {\left(\ln x\right)}^{\frac{1}{2} - 1} \left(\ln x\right) '$.............[Chain Rule]
$= \frac{1}{2 \sqrt{\ln} x} \left(\frac{1}{x}\right) = \frac{1}{2 x \sqrt{\ln} x} .$

$f ' \left(4\right) = \frac{1}{8 \sqrt{\ln} 4} .$

Recall that$f ' \left(4\right)$ is the slope of tgt. line to $f$ at $x = 4.$

Normal is perp. to tgt., hence, slope of normal $= - \frac{1}{f ' \left(4\right)} = - 8 \sqrt{\ln} 4.$

It passes thro. pt. $\left(4 , f \left(4\right)\right) = \left(4 , \sqrt{\ln} 4\right)$

$\therefore$ The eqn. of normal is, $y - \sqrt{\ln} 4 = - \left(8 \sqrt{\ln} 4\right) \left(x - 4\right) ,$ written in std. form, $\left(8 \sqrt{\ln} 4\right) x + y - 32 \sqrt{\ln} 4 - \sqrt{\ln} 4 = 0 , .$ i.e., $\left(8 \sqrt{\ln} 4\right) x + y - 33 \sqrt{\ln} 4 = 0.$