How do you differentiate #f(x) = xe^(-2x^2)# using the product rule?

1 Answer
Nov 14, 2015

Answer:

#f'(x)=(1-4x^2)/(e^(2x^2))#

Explanation:

We have two terms: #x# and #e^(-2x^2)#. The product rule states that the derivative of the product of the two will equal the following:
#f'(x)=color(red)(d/(dx)[x]*e^(-2x^2))+color(blue)(x*d/(dx)[e^(-2x^2)])#

Let's take a moment to calculate both of the derivatives inside of our product rule statement.
#color(red)(d/(dx)[x]=1)#
The next derivative will require the chain rule. Also, remember that the derivative of #e^x# is #e^x#.
#color(blue)(d/(dx)[e^(-2x^2)]=d/(dx)[-2x^2]*e^(-2x^2)=-4x(e^(-2x^2))#

Now that we know what both our derivatives are equal to, let's plug them back into our equation.

#f'(x)=color(red)(1*e^(-2x^2))+color(blue)(x*-4x(e^(-2x^2)))#
#f'(x)=e^(-2x^2)-4x^(2)(e^(-2x^2))#
#f'(x)=e^(-2x^2)(1-4x^2)#
#color(green)(f'(x)=(1-4x^2)/(e^(2x^2)))#