If y=(x^2-4)^5(3x+4)^4
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Answer:
[tex]y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)[/tex]
Step-by-step explanation:
[tex]y=(x^2-4)^5(3x+4)^4[/tex]
[tex]y'=[\frac{d}{dx}(x^2-4)^5](3x+4)^4+(x^2-4)^5[\frac{d}{dx}(3x+4)^4][/tex]
[tex]y'=10x(x^2-4)^4(3x+4)^4+(x^2-4)^5(12(3x+4)^3)[/tex]
[tex]y'=10x(x^2-4)^4(3x+4)^4+12(x^2-4)^5(3x+4)^3[/tex]
[tex]y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)[/tex]