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

Question

Paisley Johnson · Accepted Answer

#2e^x + x + (x-2)(2e^x + 1) #

Differentiate using the #color(blue)" product rule " #

f'(x) = g(x).h'(x) + h(x).g'(x) is the result if f(x) = g(x).h(x).

the standard derivative : #d/dx(e^x) = e^x # #" ----------------------------------------------------"#

g(x) = # 2e^x + x rArr g'(x) = 2e^x + 1 #

and h(x) = x - 2 # rArr h'(x) = 1 # #" -------------------------------------------------"#

Changing these numbers in f'(x)

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

Samantha Adkison · Answer

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

1. Identify the two functions: $ u(x) = 2e^x + x $ and $ v(x) = x - 2 $.
2. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
3. Find the derivatives:
   - $ u'(x) $ is the derivative of $ u(x) $, which is $ (2e^x + 1) $.
   - $ v'(x) $ is the derivative of $ v(x) $, which is $ 1 $.
4. Substitute into the product rule formula:
   $ f'(x) = (2e^x + 1)(x - 2) + (2e^x + x)(1) $.
5. Simplify the expression:
   $ f'(x) = (2e^x + 1)(x - 2) + (2e^x + x) $.
   $ f'(x) = 2e^x(x - 2) + (x - 2) + 2e^x + x $.
6. Further simplify:
   $ f'(x) = 2xe^x - 4e^x + x - 2 + 2e^x + x $.
   $ f'(x) = 2xe^x - 2e^x + 2e^x + x - 2 + x $.
7. Combine like terms:
   $ f'(x) = 2xe^x + 2x - 2 $.