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ksaimouli
 3 years ago
dy/dx
ksaimouli
 3 years ago
dy/dx

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ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350749050460:dw

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2Just product rule term by term! Start with u = x² > u' = ...?

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350749192642:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0power rule and then e^x is e^x

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2If 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350749473100:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0i took the power rule so it is 2x and e^x is e^x

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2Then where is your PRODUCT rule for the first term?

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0hmm ok dw:1350749608453:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350749664511:dw

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2I'm lost, totally :( Now, you should follow my guiding: For the first term: u = x² > u' = v = e^x > v' = => u' v + uv' = ....

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350750048815:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350750099466:dw

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, at least some light shining out of the tunnel here :)

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350750252467:dw

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2Make 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
 3 years ago
Best ResponseYou've already chosen the best response.0i used the same formula for this one but i did not get

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350750484771:dw

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2Does it look simple enough?

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350750529693:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0this is wrong but i dont know where i did the mistake

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2So don't make it complicated by jumping all over, make each line is worthy effort :)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1x^2(e^x)  x(e^x) ... make your life simple, pull out e^x

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1e^x * (x^2  x), now use the product rule

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1e^x * (2x1) + (x^2  x) * e^x

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Now simplify, pull out e^x again: e^x (2x  1 + x^2  x) = e^x (x^2 + x  1)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0what about this one dw:1350751090719:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0if i take the product rule the e^2

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350751424118:dw

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.0can u plz help with this
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