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Dido525 Group Title

Find the derivative:

  • 2 years ago
  • 2 years ago

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  1. Dido525 Group Title
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    |dw:1349927470125:dw|

    • 2 years ago
  2. Dido525 Group Title
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    I can easily find the derivative of the ln(x+ln(x+ln(x))) but I cant figure out how to find the derivative of the first part.

    • 2 years ago
  3. sritama Group Title
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    i think it will be better to take the log of that part with the respect of e base.

    • 2 years ago
  4. Dido525 Group Title
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    |dw:1349927833215:dw|

    • 2 years ago
  5. sritama Group Title
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    |dw:1349927846691:dw|

    • 2 years ago
  6. sritama Group Title
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    yeah.

    • 2 years ago
  7. Dido525 Group Title
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    Okay I had an idea about that. But don't you have to do that for the ln(x+ln(x+ln(x))) as well?

    • 2 years ago
  8. Dido525 Group Title
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    Like I thought you had to take the log of both sides for logarithmic differentiation.

    • 2 years ago
  9. sritama Group Title
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    yes,or you may solve the two parts one by one

    • 2 years ago
  10. Dido525 Group Title
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    So I thought you had to:|dw:1349927977970:dw|

    • 2 years ago
  11. Dido525 Group Title
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    Do you do that?

    • 2 years ago
  12. Dido525 Group Title
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    Don't*

    • 2 years ago
  13. sritama Group Title
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    and even if you take the ln of ln(x+ln(x+ln(x))),there ain't any problem

    • 2 years ago
  14. Dido525 Group Title
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    Yes yes. I knew that. But is that valid? I though you had to Log both sides of the ENTIRE term.

    • 2 years ago
  15. Dido525 Group Title
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    Ohh? Why not?

    • 2 years ago
  16. sritama Group Title
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    yes it is.

    • 2 years ago
  17. Dido525 Group Title
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    How is it valid? Sorry for being stubborn.

    • 2 years ago
  18. hartnn Group Title
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    sorry to interrupt. but ln(A+B)\(\ne\)ln A + ln B so here you have to differentiate 1st term separately , using ln

    • 2 years ago
  19. Dido525 Group Title
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    Ohh I know @hartnn . I didn't seperate it.

    • 2 years ago
  20. sritama Group Title
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    |dw:1349928327564:dw|

    • 2 years ago
  21. sritama Group Title
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    that is why you can find the derivatives separately

    • 2 years ago
  22. Dido525 Group Title
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    Lol I know. By I mean why can you ln only the (arcsin)^Tan(x)? I though you had to ln EVERYTHING.

    • 2 years ago
  23. hartnn Group Title
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    "had to" <---no

    • 2 years ago
  24. sritama Group Title
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    why?? if you solve them part by part, you are free to take the ln of one part without changing the other

    • 2 years ago
  25. Dido525 Group Title
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    Ohh. Okay. That explains it.

    • 2 years ago
  26. Dido525 Group Title
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    Fair enough haha...

    • 2 years ago
  27. Dido525 Group Title
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    Could I do this instead? |dw:1349928628192:dw|

    • 2 years ago
  28. Dido525 Group Title
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    For the derivative I mean.

    • 2 years ago
  29. Dido525 Group Title
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    @sritama ,@hartnn

    • 2 years ago
  30. Dido525 Group Title
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    @hartnn

    • 2 years ago
  31. hartnn Group Title
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    NO! what have u done!!

    • 2 years ago
  32. sritama Group Title
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    i couldn't get it @Dido525 :(

    • 2 years ago
  33. Dido525 Group Title
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    You make me chukle @hartnn .

    • 2 years ago
  34. sritama Group Title
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    how and why will you do that?

    • 2 years ago
  35. Dido525 Group Title
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    Well the derivative of a^x = a^x *ln(a)

    • 2 years ago
  36. hartnn Group Title
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    ln y = tan x ln(sin^-1 (x)) diff. this. no other way

    • 2 years ago
  37. Dido525 Group Title
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    Okay okay :P .

    • 2 years ago
  38. Dido525 Group Title
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    Then this question isn't so hard.

    • 2 years ago
  39. hartnn Group Title
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    is it ? what u get as y' ?

    • 2 years ago
  40. Dido525 Group Title
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    Wait.

    • 2 years ago
  41. Dido525 Group Title
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    Doing it now.

    • 2 years ago
  42. Dido525 Group Title
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    I got: |dw:1349929346410:dw|

    • 2 years ago
  43. Dido525 Group Title
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    For the derivative of (sin-1(x))^(tan(x))

    • 2 years ago
  44. Dido525 Group Title
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    @hartnn

    • 2 years ago
  45. hartnn Group Title
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    seems correct.

    • 2 years ago
  46. Dido525 Group Title
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    Okay :D . Thanks!

    • 2 years ago
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