Product Derivative Example

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Jason76

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\(\displaystyle y = (2x - 2)^{4}(x^{2} + x + 1)^{5}\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][\dfrac{d}{dx}(2x - 2)^{4}] + [(2x - 2)^{4}][\dfrac{d}{dx}(x^{2} + x + 1)^{5} ]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} du] + [(2x - 2)^{4}][5(v)^{4} dv]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} (2)] + [(2x - 2)^{4}][5(v)^{4} (2x^{3} + 1)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][8 (u)^{3} ] + [(2x - 2)^{4}][(v)^{4} (10x^{3} + 5)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][(2x - 2)^{3} (8) ] + [(2x - 2)^{4}][(x^{2} + x + 1)^{4} (10x^{3} + 5)]\) :confused:
 
Last edited:
Jason76 said:
\(\displaystyle y = (2x - 2)^{4}(x^{2} + x + 1)^{5}\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][\dfrac{d}{dx}(2x - 2)^{4}] + [(2x - 2)^{4}][\dfrac{d}{dx}(x^{2} + x + 1)^{5} ]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} du] + [(2x - 2)^{4}][5(v)^{4} dv]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} (2)] + [(2x - 2)^{4}][5(v)^{4} (2x^{3} + 1)] \ \ \ \ \)<----No, the derivative of \(\displaystyle \ \ x^2 + x + 1 \ \ \ is \ \ \ 2x + 1, \ \ \ not \ \ \ 2x^3 + 1.\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][8 (u)^{3} ] + [(2x - 2)^{4}][(v)^{4} (10x^{3} + 5)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][(2x - 2)^{3} (8) ] + [(2x - 2)^{4}][(x^{2} + x + 1)^{4} (10x^{3} + 5)]\) :confused:
Click to expand...
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\(\displaystyle y = (2x - 2)^{4}(x^{2} + x + 1)^{5}\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][\dfrac{d}{dx}(2x - 2)^{4}] + [(2x - 2)^{4}][\dfrac{d}{dx}(x^{2} + x + 1)^{5} ]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} du] + [(2x - 2)^{4}][5(v)^{4} dv]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} (2)] + [(2x - 2)^{4}][5(v)^{4} (2x + 1)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][(u)^{3} (8)] + [(2x - 2)^{4}][(v)^{4} (5x + 5)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][(2x - 2)^{3} (8)] + [(2x - 2)^{4}][(x^{2} + x + 1)^{4} (5x + 5)]\)
 
Still getting wrong answer on computer homework. What could be wrong? :confused:
 
Below is the correct answer that the online homework wanted:

\(\displaystyle y = (2x - 2)^{4}(x^{2} + x + 1)^{5}\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][\dfrac{d}{dx}(2x - 2)^{4}] + [(2x - 2)^{4}][\dfrac{d}{dx}(x^{2} + x + 1)^{5} ]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} du] + [(2x - 2)^{4}][5(v)^{4} dv]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][4(u)^{3} (2)] + [(2x - 2)^{4}][5(v)^{4} (2x + 1)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][(u)^{3} (8)] + [(2x - 2)^{4}][(v)^{4} (10x + 5)]\)

\(\displaystyle y' = [(x^{2} + x + 1)^{5}][(2x - 2)^{3} (8)] + [(2x - 2)^{4}][(x^{2} + x + 1)^{4} (10x + 5)]\)
 
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