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Author Question: Find the equation of the line tangent to the curve y = x ln x2 at the point (Read 358 times)

geoffrey

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on: Oct 21, 2019
Find the equation of the line tangent to the curve y = x ln x2 at the point
◦ y = 2x + 1
◦ y = 3x - 3
◦ y = 2x - 2
◦ y = x - 1
Basic Technical Mathematics with Calculus ¦ Washington ¦ 10th Edition
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Marked as best answer by geoffrey on Oct 21, 2019

gcook

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Reply #1 on: Oct 21, 2019
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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geoffrey

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Reply 2 on: Oct 21, 2019
Great answer, keep it coming :)

komodo7

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Reply 3 on: Yesterday
YES! Correct, THANKS for helping me on my review

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