# What is the slope of the line normal to the tangent line of #f(x) = xe^(x-3)+x^3 # at # x= 2 #?

##### 1 Answer

#### Explanation:

Before anything else, find the function's derivative. To do this, we will need the **product rule** for the **power rule**.

#f'(x)=e^(x-3)d/dx(x)+xd/dx(e^(x-3))+3x^2#

The simpler derivative here is **chain rule**.

Since

Thus,

Plugging this back into our

#f'(x)=e^(x-3)(1)+x(e^(x-3))+3x^2#

#f'(x)=e^(x-3)(x+1)+3x^2#

Now, we must find the slope of the tangent line, which is equal to the value of the derivative at

#f'(2)=e^(2-3)(2+1)+3(2^2)=e^-1(3)+3(4)=3/e+12#

This, however, is the slope of the tangent line. Since perpendicular lines have opposite reciprocal slopes, the slope of the normal line is

#-1/(3/e+12)=-1/((3+12e)/e)=-e/(3+12e)#