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Find the derivative of $(5x^3+3x-1)(x-1)$

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Let $f(x) = (5x^3+3x-1)(x-1)$
By Leibnitz product rule,
$ f'(x) = (5x^3+3x-1) \large\frac{d}{dx}$$(x-1)+(x-1) \large\frac{d}{dx}$$(5x^3+3x-1) $
$ = (5x^3+3x-1) (1) +(x-1)(5.3x^2+3-0)$
$ = (5x^3+3x-1) +(x-1)(15x^2+3)$
$ = 5x^3+3x-1+15x^3+3x-15x^2-3$
$ = 20x^3-15x^2+6x-4$
answered Apr 9, 2014 by thanvigandhi_1

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