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

Question

Ezekiel Alexander · Accepted Answer

See the explanation below.

Let #y = x-xe^(x-x^2/2)#. Factor out #x#. #y = x(1-e^(x-x^2/2))# Take the derivative and use the #color(blue)("product rule")# : #dy/dx = color(red)(d/dx(x))(1-e^(x-x^2/2)) + xcolor(red)(d/dx (1-e^(x-x^2))# The #1# doesn't matter when differentiating. Let #g(x) = e^(x-x^2/2)# Using the property that #color(blue)(d/dx -f(x) = -d/dx f(x))#, we get : #dy/dx = 1-e^(x-x^2/2)-x*color(red)(g'(x)# We have to solve the for #color(red)(g'(x))#. Take the natural logarithm of #g(x)# : #ln(g(x)) = (x-x^2/2)ln e = x-x^2/2# And now differentiate both sides : #(g'(x))/(g(x)) = d/dx x-x^2/2 = d/dx x - d/dx x^2/2# #(g'(x))/(g(x)) = 1 - x# #g'(x) = (e^(x-x^2/2))(1-x)# Substitute it in the #dy/dx# equation : #dy/dx = 1 -e^(x-x^2/2) - x(e^(x-x^2/2))(1-x)#, so #dy/dx = 1-e^(x-x^2/2)(1-x(1-x))# #dy/dx = 1-e^(x-x^2/2)(1-x+x^2)# And finally, #color(blue)(f'(x) =1-e^(x-x^2/2)(x^2+x-1) #.

Isabella Ackerman · Answer

undefined To differentiate $ f(x) = x - xe^{x - \frac{x^2}{2}} $ using the product rule, follow these steps:

1. Identify the functions $ u(x) $ and $ v(x) $ such that $ f(x) = u(x) \cdot v(x) $.
2. Differentiate $ u(x) $ to get $ u'(x) $.
3. Differentiate $ v(x) $ to get $ v'(x) $.
4. Apply the product rule formula $ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) $ to find $ f'(x) $.