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

Question

Evelyn Agard · Accepted Answer

#f'(x)=(-e^x+secxtanx)(3x^2-2x)+(-e^x+secx)(6x-2)# #f(x)=(-e^x+secx)(3x^2-2x)=g(x)h(x)# #f'(x)=g'(x)h(x)+h'(x)g(x)# #g(x)=-e^x+secx# #g'(x)=-e^x+secxtanx# #h(x)=3x^2-2x# #h'(x)=6x-2# #f'(x)=(-e^x+secxtanx)(3x^2-2x)+(-e^x+secx)(6x-2)#

Isabella Ackerman · Answer

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

1. Identify the functions $ u(x) $ and $ v(x) $.
   Let $ u(x) = -e^x + \sec(x) $ and $ v(x) = 3x^2 - 2x $.

2. Compute the derivatives of $ u(x) $ and $ v(x) $.
   $ u'(x) = -e^x $ (derivative of $-e^x$) and $ v'(x) = 6x - 2 $ (derivative of $ 3x^2 - 2x $).

3. Apply the product rule:
   $ f'(x) = u'(x)v(x) + u(x)v'(x) $.

4. Substitute the derivatives and the original functions:
   $ f'(x) = (-e^x)(3x^2 - 2x) + (-e^x + \sec(x))(6x - 2) $.

5. Simplify the expression:
   $ f'(x) = -3x^2e^x + 2xe^x + 6x\sec(x) - 2\sec(x) $.

Thus, the derivative of $ f(x) $ with respect to $ x $ using the product rule is $ -3x^2e^x + 2xe^x + 6x\sec(x) - 2\sec(x) $.