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DLS

  • 2 years ago

DE help

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  1. DLS
    • 2 years ago
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    \[y=\log(x^{2}\sqrt{x^{2}+1})\]

  2. DLS
    • 2 years ago
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    I got something like this \[\frac{1}{x^{2}\sqrt{x^{2}+1}} \times 2\sqrt{x^{2}+1} \times 2x\]

  3. DLS
    • 2 years ago
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    @ghazi

  4. ghazi
    • 2 years ago
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    there is a bit of editing you should have used product rule

  5. ghazi
    • 2 years ago
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    for (x^2 sqrt{x^2+1})

  6. DLS
    • 2 years ago
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    how?:o

  7. ghazi
    • 2 years ago
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    \[\frac{ d(x^2 \sqrt{x^2+1}) }{ dx }=x^2\frac{ d \sqrt{x^2+1} }{ dx }+\sqrt{x^2+1}*2x\]

  8. DLS
    • 2 years ago
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    woah :o thanks

  9. ghazi
    • 2 years ago
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    :) YW

  10. DLS
    • 2 years ago
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    how did you get dx in the denominator?

  11. ghazi
    • 2 years ago
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    you are differentiating with respect to x

  12. ghazi
    • 2 years ago
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    i can write it as\[\frac{ d }{ dx }(\sqrt{x^2+1}*x^2)\]

  13. DLS
    • 2 years ago
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    oh wait equation is messed up lol just asec

  14. DLS
    • 2 years ago
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    \[\frac{dy}{dx}=\frac{dlog(x^{2}\sqrt{x^{2}+1}}{dx^{2}\sqrt{x^{2}+1}} \times \frac{dx^{2} \sqrt{x^{2}+1}}{d \sqrt{x^{2}+1}} \] ..so on

  15. ghazi
    • 2 years ago
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    NO

  16. DLS
    • 2 years ago
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    i do like this,where i didnt diff. wrt. dx

  17. DLS
    • 2 years ago
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    thats how im taught >.>

  18. ghazi
    • 2 years ago
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    you have to differentiate wrt x, you can't diffrentiate wrt to function

  19. DLS
    • 2 years ago
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    wth :/ am i taught wrong :S but i was getting all the answers like this

  20. ghazi
    • 2 years ago
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    differentiation is done wrt to a variable basically differentiation tells us rate of change of one variable wrt to another and function is something that tells us the relation between those two variables so we can never get rate of change wrt to a function

  21. DLS
    • 2 years ago
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    okay..!

  22. ghazi
    • 2 years ago
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    \[\frac{ d }{ dx }(\sqrt{x^2+1})*x^2+2x* \sqrt{x^2+1}\]

  23. DLS
    • 2 years ago
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    okay...clear!

  24. ghazi
    • 2 years ago
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    cool :D

  25. ghazi
    • 2 years ago
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    \[\frac{ d }{ dx } (\sqrt{x^2+1})=\frac{ 1 }{ \sqrt{x^2+1} }*\frac{ d }{ dx }(x^2+1)\]

  26. ghazi
    • 2 years ago
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    there will be a factor of -1/2 multiplied in RHS

  27. DLS
    • 2 years ago
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    derivative of sqrt x is 1/2x right :o

  28. ghazi
    • 2 years ago
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    -1/2 x

  29. DLS
    • 2 years ago
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    \[\frac{d \sqrt{x}}{dx}= \frac{1}{2 \sqrt{x}}\] does my coaching class suck lol :/

  30. DLS
    • 2 years ago
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    http://wiki.answers.com/Q/What_is_the_derivative_of_square_root_of_x

  31. ghazi
    • 2 years ago
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    no they are right let me show you

  32. DLS
    • 2 years ago
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    will it be like this: \[\frac{1}{2 \sqrt{x^{2}+1}} \times 2x = \frac{x}{\sqrt{x^{2}+1}} \]

  33. ghazi
    • 2 years ago
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    \[\frac{ d }{ dx } (x)^{-1/2}= \frac{ -1 }{ 2 }(x)^{-1/2-1}\]

  34. DLS
    • 2 years ago
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    so how do we have 2 answers

  35. ghazi
    • 2 years ago
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    no we have one, just be careful in calculating power

  36. DLS
    • 2 years ago
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    okay..

  37. DLS
    • 2 years ago
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    so final answer im getting is

  38. DLS
    • 2 years ago
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    \[\frac{1}{x(x^{2}+1}\]

  39. DLS
    • 2 years ago
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    \[\frac{1}{x^{2} \sqrt{x^{2}+1}} \times \frac{x}{\sqrt{x^{2}+1}}\]

  40. DLS
    • 2 years ago
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    which is wrong :/ http://www.wolframalpha.com/input/?i=derivative+of+log%28x%5E2%28sqrt%28x%5E2%2B1%29%29%29

  41. ghazi
    • 2 years ago
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    \[\frac{ d }{ dx } (\sqrt{x^2+1})=\frac{1 }{ 2 }(\sqrt{x^2+1})^{-3/2}* 2x\] your answer

  42. DLS
    • 2 years ago
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    what did you do

  43. ghazi
    • 2 years ago
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    \[\frac{ d }{ dx }(x^n)=nx^{n-1}\] hope you know this formula

  44. DLS
    • 2 years ago
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    oh okay!

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