How do you differentiate #g(y) =(2x^3+x )e^x # using the product rule?

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Answer 1 · 2018-04-22T18:00:14

If #g(y)#: #g'(y) = 0#

If #g(x)#: #g'(x) = (2x^3+7x+1) e^x#

#g(y) = (2x^3 +x)e^x#

#g'(y) = 0#

If you meant #g(x) = (2x^3+x)e^x#, then

#g'(x) = d/(dx)[(2x^3+x)]e^x + (2x^3+x)d/(dx)[e^x]#

#g'(x) = (6x+1)e^x + (2x^3+x)e^x#

#g'(x) = (6x+1 + 2x^3 + x)e^x#

#g'(x) = (2x^3+7x+1) e^x#