How do you differentiate #f(x)=(4e^x+1)(x^2-3)# using the product rule?

Question

Ezra Adams · Accepted Answer

#f'(x) = 4e^x(x^2+2x-13) + 2x# #f(x) = (4e^x+1)(x^2-3)# Click here for details on the product rule. #f'(x) = d/dx(4e^x + 1)(x^2-3) + (4e^x + 1)d/dx(x^2-3)# #f'(x) = (4e^x)(x^2-3) + (4e^x + 1)(2x)# To simplify, we'll expand and see if we can condense it later: #f'(x) = 4 e^x x^2 - 12e^x + 8 x e^x + 2x# #f'(x) = 4e^x(x^2+2x-13) + 2x#

Evelyn Agard · Answer

undefined To differentiate $ f(x) = (4e^x + 1)(x^2 - 3) $ using the product rule, you can follow these steps:

1. Identify the two functions being multiplied: $ u(x) = 4e^x + 1 $ and $ v(x) = x^2 - 3 $.
2. Apply the product rule formula: $ (uv)' = u'v + uv' $.
3. Find the derivatives of $ u(x) $ and $ v(x) $:
   $ u'(x) = 4e^x $ and $ v'(x) = 2x $.
4. Plug these derivatives into the product rule formula:
   $ f'(x) = (4e^x + 1)(2x) + (4e^x + 1)(x^2 - 3) $.
5. Simplify the expression:
   $ f'(x) = (8xe^x + 2x) + (4e^x(x^2 - 3) + (x^2 - 3)) $.
6. Further simplify if necessary:
   $ f'(x) = 8xe^x + 2x + 4xe^x - 12e^x + x^2 - 3 $.
7. Combine like terms to obtain the final result:
   $ f'(x) = 8xe^x + 4xe^x + 2x - 12e^x + x^2 - 3 $.