swanny 3 years ago derivative of (x^(2))(e^(1/x))

1. swanny

@Callisto

2. swanny

i got (2x)(e^(1/x))+x^(2)(e^(1/x) which is wrong

3. satellite73

yeah it is

4. satellite73

$g(x)=e^{\frac{1}{x}}$ $g'(x)=-\frac{1}{x^2}e^{\frac{1}{x}}$ by the chain rule

5. satellite73

so the second part of your product rule is the problem. first part is good

6. swanny

but how is it (-1/x^2) e^(1/x)?

7. swanny

shouldn't the derivative e^(1/x)

8. Callisto

$y=e^{\frac{1}{x}}$ Let u = 1/x $\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?$

9. swanny

1?

10. swanny

or 0

11. swanny

by quotient rule i did (1)'(x)-(1)(x)'/(x)^@

12. Callisto

You can use power for d/dx (1/x) $\frac{d}{dx} \frac{1}{x} = \frac{d}{dx} (x^{-1}) = (-1)x^{-1-1} =...?$

13. swanny

-x^(-2)?

14. Callisto

Yes.

15. swanny

ok so it will be e^(-x^(-2))?

16. Callisto

The power of e would not change when you differentiate e^(something)!

17. Callisto

$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{d}{du}e^u \times \frac{d}{dx}(\frac{1}{x})= ...?$ u = 1/x and you found d/dx(1/x). So......

18. swanny

i'm lost with this one

19. Callisto

Where?

20. swanny

are you asking for the derivative of (1/x) above? i dont understand what exactly you are asking

21. Callisto

I asked derivative of 1/x because you didn't get that right. To get the derivative of derivative of (x^(2))(e^(1/x)), it should be like this: $y = x^2e^{\frac{1}{x}}$$y' = x^2\frac{d}{dx}e^{\frac{1}{x}} + e^{\frac{1}{x}}\frac{d}{dx}(x^2)$ Then, then you need to work out what d/dx (e^(1/x)) is and d/dx (x^2) are. The later one is easy, and you got that right. The problem is to find d/dx (e^(1/x)) So, let u = 1/x $\frac{d}{dx}e^{\frac{1}{x}} = \frac{d}{du}(e^u) \times \frac{d}{dx} (\frac{1}{x}) = ...?$

22. Callisto

Is that clear? Do you understand what we are working on?

23. swanny

but isn't derivative of (1/x) = -x^(-2) ? i thought we got that right

24. swanny

wait. give me second

25. Callisto

Yes. d/dx (1/x) = -x^(-2) = -1/x^2

26. swanny

so you are asking derivative of e^(1/x) * -1/x^2?

27. Callisto

No. Instead, derivative of e^(1/x), which is equal to e^(1/x) * -1/x^2. Do you understand how to get it?

28. swanny

so looking at the original question question my answer should be (2x)(e^1/x)-(x^2)(e^(1/x))?

29. swanny

derivative of (x^(2))(e^(1/x)) is the original question

30. Callisto

No... The second part is not correct. For the second part, you need to find the derivative of e^(1/x). What is it?

31. swanny

isn't it e^(1/x)

32. swanny

e^(1/x) * -1/x^2?

33. Callisto

It is *NOT* e^(1/x) e^(1/x) * -1/x^2 ^How do you get it?

34. swanny

because derivative of 1/x = -1/x^2

35. Callisto

Yup.. So, can you solve $$y' = x^2\frac{d}{dx}e^{\frac{1}{x}} + e^{\frac{1}{x}}\frac{d}{dx}(x^2)$$ now?

36. Callisto

37. swanny

yes i had already gotten the first part. just the second was the problem

38. Callisto

Just show us what you've got.

39. swanny

final answer for the original question is (2x)(e^(1/x))-(x^2)(1/x^2)(e^(1/x))

40. swanny

this is what i got

41. Callisto

That looks pretty cool~ But you can simplify the last term.

42. swanny

i don't think we have to. but you can show me if you want

43. Callisto

$(2x)(e^{\frac{1}{x}})-(x^2)\frac{1}{x^2}(e^{\frac{1}{x}}) = (2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =...?$

44. swanny

x^2 / x^2 gets cancelled out = 1

45. Callisto

Yes.

46. swanny

perfect.

47. swanny

Thank you very much for your help

48. swanny

you explain concepts very well

49. Callisto

You're welcome. I hope I didn't confuse you :S $(2x)(e^{\frac{1}{x}})-\frac{x^2}{x^2}(e^{\frac{1}{x}}) =(2x)(e^{\frac{1}{x}})-e^{\frac{1}{x}}=e^{\frac{1}{x}}(2x-1)$That looks nice :)

50. swanny

No, not confusing at all

51. swanny

yes that looks better

52. Callisto

:)