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Author Question: Given that the relation 3x2 + dt = 3 defines y implicitly as a differentiable function of x,find . (Read 22 times)

asd123

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on: May 26, 2021

Question 1

Find the derivative of F(x) = dt.
◦ 2e2x cos (e2x)
◦ ex cos (e4x)
◦ 2e2x sin (e4x)
◦ e2x cos (e4x)
◦ 2e2x cos (e4x)

Question 2

Given that the relation  3x2 + dt = 3  defines y implicitly as a differentiable function of x,
find .
◦ 6x +cos(y) - ysin (y)
◦
◦ 6x + cos (t) - t sin (t)
◦
◦ 6x + ycos (y)
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Marked as best answer by asd123 on May 26, 2021

sarahccccc

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Reply #1 on: May 26, 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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asd123

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Reply 2 on: May 26, 2021
Thanks for the timely response, appreciate it

Jossy

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Reply 3 on: Yesterday
:D TYSM

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