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ksaimouli
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dw:1350749050460:dw

Chlorophyll
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Just product rule term by term! Start with
u = x² > u' = ...?

ksaimouli
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dw:1350749192642:dw

ksaimouli
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?

ksaimouli
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power rule and then e^x is e^x

ksaimouli
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i am not sure

ksaimouli
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@Chlorophyll

Chlorophyll
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If you're unsure, so please don't jump ahead!
You have your choices: following my guidance by fill in the blanks or show your work completely :)

ksaimouli
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dw:1350749473100:dw

ksaimouli
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i took the power rule so it is 2x and e^x is e^x

Chlorophyll
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Then where is your PRODUCT rule for the first term?

ksaimouli
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hmm ok dw:1350749608453:dw

ksaimouli
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dw:1350749664511:dw

ksaimouli
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is this correct

Chlorophyll
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I'm lost, totally :( Now, you should follow my guiding:
For the first term:
u = x² > u' =
v = e^x > v' =
=> u' v + uv' = ....

ksaimouli
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dw:1350750048815:dw

ksaimouli
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dw:1350750099466:dw

Chlorophyll
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Yes, at least some light shining out of the tunnel here :)

ksaimouli
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so whats next

ksaimouli
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dw:1350750252467:dw

ksaimouli
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so this is answer

Chlorophyll
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Make it neat, so you can easily simplify them:
( x²e^x + 2x e^x )  ( x e^x + e^x)
= e^x ( x² + 2x  x  1)
= e^x ( x² + x  1)

ksaimouli
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i used the same formula for this one but i did not get

ksaimouli
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dw:1350750484771:dw

Chlorophyll
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Does it look simple enough?

ksaimouli
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yup

ksaimouli
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dw:1350750529693:dw

ksaimouli
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this is wrong but i dont know where i did the mistake

Chlorophyll
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So don't make it complicated by jumping all over, make each line is worthy effort :)

bahrom7893
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x^2(e^x)  x(e^x) ... make your life simple, pull out e^x

bahrom7893
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e^x * (x^2  x), now use the product rule

bahrom7893
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e^x * (2x1) + (x^2  x) * e^x

bahrom7893
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Now simplify, pull out e^x again:
e^x (2x  1 + x^2  x) = e^x (x^2 + x  1)

bahrom7893
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Now, you can either leave it like that, or distribute e^x. If you choose to do the latter,
e^x * x^2 + e^x * x  e^x

ksaimouli
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what about this one dw:1350751090719:dw

ksaimouli
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@Chlorophyll

ksaimouli
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if i take the product rule the e^2

ksaimouli
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dw:1350751424118:dw

ksaimouli
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can u plz help with this