If y=(x^2-4)^5(3x+4)^4

Chiseled asked Jul 18, 2023

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Chiseled asked Jul 18, 2023

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17 votes

Answer:

y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)

Explanation:

y=(x^2-4)^5(3x+4)^4

y'=[(d)/(dx)(x^2-4)^5](3x+4)^4+(x^2-4)^5[(d)/(dx)(3x+4)^4]

y'=10x(x^2-4)^4(3x+4)^4+(x^2-4)^5(12(3x+4)^3)

y'=10x(x^2-4)^4(3x+4)^4+12(x^2-4)^5(3x+4)^3

y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)

Tbthorpe answered Jul 25, 2023

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