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

##### 1 Answer

#### Explanation:

First, find the point the normal line will intercept.

#f(3)=3^ee^3#

The normal line will pass through the point

To find the slope of the normal line, we should first find the derivative of the function. This can be found through the product rule:

#f'(x)=e^xd/dx[x^e]+x^ed/dx[e^x]#

Note that

Thus,

#f'(x)=e^x(ex^(e-1))+x^ee^x#

#f'(x)=e^x(ex^(e-1)+x^e)#

The slope of the *tangent* line at

#f'(3)=e^3(e(3^(e-1))+3^e)#

#f'(3)=e^3 3^(e-1)(e+3)#

However, we want to find the *normal* line of the function. The slopes of the tangent line and normal line are perependicular, to they are opposite reciprocals. The opposite reciprocal of

#-(e^3 3^(e-1)(e+3))^-1=-(3^(1-e))/(e^3(e+3))#

The slope of

#y-3^ee^3=-(3^(1-e))/(e^3(e+3))(x-3)#