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Author Question: Differentiate.f(x) = (6x5- 3x3- 4)(3x2- 3)[Do not use algebra before differentiating] (Read 68 times)

NClaborn

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on: Jun 16, 2021
Differentiate.

f(x) = (6x5 - 3x3 - 4)(3x2 - 3)
[Do not use algebra before differentiating]
◦ f'(x) = (3x2 - 3)(30x4 - 9x2 - 4) + (6x5 - 3x3 - 4)
◦ f'(x) = (3x2 - 3)(6x4 - 3x2) + (6x5 - 3x3 - 4)
◦ f'(x) = (3x2 - 3)(30x4 - 9x2) + (6x5 - 3x3 - 4)
◦ f'(x) = (30x4 - 9x2)
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Marked as best answer by NClaborn on Jun 16, 2021

lkoler

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Reply #1 on: Jun 16, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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NClaborn

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Reply 2 on: Jun 16, 2021
YES! Correct, THANKS for helping me on my review

samiel-sayed

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Reply 3 on: Yesterday
Excellent

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