How do you differentiate #g(y) =e^x(x^3-1) # using the product rule?

Question

1330 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-27T18:14:55

#g'(y)=0#

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

The product rule states that

#d/dx[f(x)*g(x)]=f(x)*g'(x)+g(x)*f'(x)#.

Application of this rule in this particular function yields

Case 1 - g(y) as written in the question

The function only contains x terms and so the derivative #g'(y)=0#.

Case 2 - Assuming g(x) as then the product rule becomes applicable

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

#=e^x(3x^2+x^3-1)#