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

First, find the point the normal line will intercept.

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

The normal line will pass through the point #(1,e)#.

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 #x=1#, we can take the opposite reciprocal of that to find the slope of the normal line, since the normal line is perpendicular to the tangent line and perpendicular lines have opposite reciprocal slopes.

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 #0#. Thus, the normal line, which is perpendicular, will have an undefined slope.

The vertical line that passes through the point #(1,e)# is #x=1#.

Graphed are the function and its normal line: