F(x)=(4t^2-t) (t^3-8t^2+12) in product rule.

1 answer:

6 0

$\dfrac{d}{dt} [(4t^2-t)(t^3 -8t^2+12)]\\\\\\=(4t^2-t)\dfrac{d}{dt}(t^3-8t^2 +12) + (t^3 -8t^2 +12) \dfrac{d}{dt} (4t^2 -t)\\\\\\=(4t^2 -t)(3t^2-16t)+(t^3-8t^2+12)(8t-1)\\\\=12t^4-64t^3 -3t^3+16t^2+8t^4-64t^3+96t-t^3+8t^2-12\\\\=20t^4-132t^3+24t^2+96t-12$

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