How do you differentiate #g(y) =(x^3 + x)(3x-5) # using the product rule?

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How do you differentiate #g(y) =(x^3 + x)(3x-5) # using the product rule?

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Answer 1 · 2017-01-19T05:45:58

#g'(y)=12x^3-15x^2+6x-5#

Explanation:

Let #u=(x^3+x)#
#d'u=3x^2+1#

let #v=(3x-5)#
#d'v=3#

#g(y)=uv#
#g'(y)=ud'v+vd'u#
#g'(y)=(x^3+x)(3)+(3x-5)(3x^2+1)#
#g'(y)=3x^3+3x+9x^3+3x-15x^2-5#
#g'(y)=3x^3+9x^3-15x^2+3x+3x-5#
#g'(y)=12x^3-15x^2+6x-5#

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How do you differentiate #g(y) =(x^3 + x)(3x-5) # using the product rule?

1 Answer

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