Finding derivative college calculus
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Finding derivative college calculus

[From: ] [author: ] [Date: 12-04-04] [Hit: ]
It can be simplified just a little bit.h(x) = [24x(3x^2 -4)^3 *(x^3 -6x+4)^7] + [(21x^2 -42)*(3x^2 -4)^4 * (x^3 -6x+4)^6]-Power rule, chain rule, product rule. Lots of them. Im not doing it for you.......
Hi, if someone can help with steps I will give best answer, I really want to get it
Find h ' (x) if h (x) = (3x^2 - 4)^4 (x^3 - 6x + 4) ^7

Thanks

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h'(x) = [4(3x^2 -4)^3 * (6x)*(x^3-6x+4)^7] + [(3x^2 -4)^4 7(x^3 -6x+4)^6 *(3x^2-6)]

It can be simplified just a little bit.

h'(x) = [24x(3x^2 -4)^3 *(x^3 -6x+4)^7] + [(21x^2 -42)*(3x^2 -4)^4 * (x^3 -6x+4)^6]

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Power rule, chain rule, product rule. Lots of them. I'm not doing it for you.

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h'(x) = [24x (3x^2 -4)^3 (x^3-6x+4)^7] + [(3x^2 -4)^4 (x^3 -6x+4)^6 (21x^2-42)]
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