# What is the equation of the normal line of #f(x)=e^x/x# at #x=1 #?

##### 1 Answer

#### Explanation:

First, find the point the normal line will intercept.

#f(1)=e^1/1=e#

The normal line will pass through the point

Now, to find the slope of the normal line, we must first find the slope of the tangent line. Thinking ahead, once we find the slope of the tangent line at

First, find the derivative of the function, will require the quotient rule.

#f'(x)=(xd/dx(e^x)-e^xd/dx(x))/x^2#

#f'(x)=(xe^x-e^x)/x^2#

#f'(x)=((x-1)e^x)/x^2#

The slope of the tangent line is

#f'(1)=((1-1)e^1)/1^2=0#

This is a rather odd case. This is a horizontal line, since it has a slope of

The vertical line that passes through the point

Graphed are the function and its normal line: