## ksaimouli 2 years ago dy/dx

1. ksaimouli

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2. Chlorophyll

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

3. ksaimouli

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4. ksaimouli

?

5. ksaimouli

power rule and then e^x is e^x

6. ksaimouli

i am not sure

7. ksaimouli

@Chlorophyll

8. Chlorophyll

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 :)

9. ksaimouli

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10. ksaimouli

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

11. Chlorophyll

Then where is your PRODUCT rule for the first term?

12. ksaimouli

hmm ok |dw:1350749608453:dw|

13. ksaimouli

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14. ksaimouli

is this correct

15. Chlorophyll

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

16. ksaimouli

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17. ksaimouli

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18. Chlorophyll

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

19. ksaimouli

so whats next

20. ksaimouli

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21. ksaimouli

22. Chlorophyll

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)

23. ksaimouli

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

24. ksaimouli

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25. Chlorophyll

Does it look simple enough?

26. ksaimouli

yup

27. ksaimouli

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28. ksaimouli

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

29. Chlorophyll

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

30. bahrom7893

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

31. bahrom7893

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

32. bahrom7893

e^x * (2x-1) + (x^2 - x) * e^x

33. bahrom7893

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

34. bahrom7893

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

35. ksaimouli

36. ksaimouli

@Chlorophyll

37. ksaimouli

if i take the product rule the e^2

38. ksaimouli

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39. ksaimouli

can u plz help with this