# What is the equation of the normal line of f(x)=xlnx at x=3 ?

Jan 1, 2016

$y - 3 \ln 3 = \frac{- 1}{\ln 3 + 1} \left(x - 3\right)$

#### Explanation:

First, find the point on $f \left(x\right)$ that the normal line will intersect.

$f \left(3\right) = 3 \ln 3$

The normal line will intersect the point $\left(3 , 3 \ln 3\right)$.

The tangent line at a point and its normal line are perpendicular. To find the slope of the tangent line, find $f ' \left(3\right)$. First, find $f ' \left(x\right)$.

The product rule will be necessary. It states that

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

Apply this to $x \ln x$:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(x\right) \cdot \ln x + \frac{d}{\mathrm{dx}} \left(\ln x\right) \cdot x$

Both of these derivatives are simple to find.

$f ' \left(x\right) = 1 \cdot \ln x + \frac{1}{x} \cdot x$
$f ' \left(x\right) = \ln x + 1$

Now, find the slope of the tangent line.

$f ' \left(3\right) = \ln 3 + 1$

Since the normal line is perpendicular to the tangent line, its slope will be the opposite reciprocal of that of the tangent line.

Thus, the slope of the normal line is $\frac{- 1}{\ln 3 + 1}$ and it passes through the point $\left(3 , 3 \ln 3\right)$.

Relate this information in the point-slope form of a line:

$y - 3 \ln 3 = \frac{- 1}{\ln 3 + 1} \left(x - 3\right)$

In decimal form (to $4$ places):

$y = - 0.4765 x + 4.7254$

Graphed:

graph{(y-xlnx)(-y-0.4765x+4.7254)=0 [-1.52, 16.26, -1.584, 7.306]}