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

Question

Ezekiel Alexander · Accepted Answer

#=3xe^x+x^2e^x+e^x-6x^2-4x#

According to the product rule,

#f(x)=g(x)*h(x)->f'(x)=g'(x)*h(x)+g(x)*h'(x)#

Next, it is:

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

#=2xe^x-4x^2+e^x-2x+x^2e^x-2x^2+xe^x-2x#

Let's add up the similar terms:

#=3xe^x+x^2e^x+e^x-6x^2-4x#

Scarlett Allison · Answer

undefined To differentiate $ f(x) = (x^2 + x)(e^x - 2x) $ using the product rule, follow these steps:

1. Identify the functions $ u(x) = x^2 + x $ and $ v(x) = e^x - 2x $.
2. Compute $ u'(x) $ and $ v'(x) $, which are the derivatives of $ u(x) $ and $ v(x) $ with respect to $ x $, respectively.
3. Apply the product rule formula: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
4. Substitute the derivatives and the original functions into the product rule formula to find $ f'(x) $.

Here are the computations:

$ u'(x) = 2x + 1 $

$ v'(x) = e^x - 2 $

Using the product rule formula:

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

Simplify the expression to get the final derivative $ f'(x) $.