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monroe17

  • 2 years ago

can someone help me step by step? find the derivative y=ln((e^(7x))/(sqrt(4x-5)))

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  1. satellite73
    • 2 years ago
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    before you begin taking the derivative, use the properties of the log to make this expression easier to differentiate

  2. satellite73
    • 2 years ago
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    it is the log of the whole thing right?

  3. monroe17
    • 2 years ago
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    yes.

  4. satellite73
    • 2 years ago
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    so first step would be, before beginning to take the derivative, rewrite as \[\ln(e^{7x})-\frac{1}{2}\ln(4x-5)\]

  5. satellite73
    • 2 years ago
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    are those steps clear?

  6. monroe17
    • 2 years ago
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    why is 1/2 there and not (4x-5)^(1/2)

  7. satellite73
    • 2 years ago
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    i used two facts \[\log(\frac{a}{b})=\log(a)-\log(b)\] and \[\log(a^n)=n\log(a)\]

  8. satellite73
    • 2 years ago
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    because \[\log(\sqrt{4x-5})=\log((4x-5)^{\frac{1}{2}})=\frac{1}{2}\log(4x-5)\]

  9. monroe17
    • 2 years ago
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    but 1/2 is outside the () and not inside ?

  10. Dido525
    • 2 years ago
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    You can use the rule like that. As long as it's insside the log.

  11. satellite73
    • 2 years ago
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    the one half comes right out front as a multiplier

  12. satellite73
    • 2 years ago
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    on other words, \(\log(\sqrt{x})=\frac{1}{2}\log(x)\)

  13. satellite73
    • 2 years ago
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    then one more step before differentiating since log and exp are inverse functions, you have \[\log(e^{7x})=7x\]

  14. monroe17
    • 2 years ago
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    okay, so do i need to use product rule for 1/2ln(4x-5)?

  15. satellite73
    • 2 years ago
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    oh no not at all

  16. satellite73
    • 2 years ago
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    \(\frac{1}{2}\) is just a constant , leave it there

  17. monroe17
    • 2 years ago
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    okay

  18. Dido525
    • 2 years ago
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    Well you could... But it's a waste of time.

  19. satellite73
    • 2 years ago
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    for example if you wanted the derivative of \(\frac{1}{2}x^3\) you do not use the product rule, you just say \(\frac{3}{2}x^2\)

  20. monroe17
    • 2 years ago
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    gotcha. the power rule^

  21. satellite73
    • 2 years ago
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    so now you have \[7x+\frac{1}{2}\ln(4x-5)\] so the only rule you need now is the chain rule for the second part, and also knowing what the derivative of the log is

  22. satellite73
    • 2 years ago
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    typo there, i meant \[7x-\frac{1}{2}\ln(4x-5)\]

  23. monroe17
    • 2 years ago
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    okay imma try it/

  24. satellite73
    • 2 years ago
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    ok let me know what you get

  25. Dido525
    • 2 years ago
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    |dw:1351998238123:dw|

  26. Dido525
    • 2 years ago
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    It will be useful when you do the ln part.

  27. monroe17
    • 2 years ago
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    (1/4x-5)*4?

  28. Dido525
    • 2 years ago
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    Good!

  29. monroe17
    • 2 years ago
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    7x-1/2*(1/4x-5)*4

  30. sirm3d
    • 2 years ago
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    oops, you forgot to differentiate 7x

  31. satellite73
    • 2 years ago
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    careful of the first term derivative of \(7x\) is just \(7\)

  32. satellite73
    • 2 years ago
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    and you might not want to write \[-\frac{4}{2(4x-5)}\] since you can cancel a 2

  33. monroe17
    • 2 years ago
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    i thought i was just doing chain rule? i know that that's 7?

  34. satellite73
    • 2 years ago
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    chain rule for \(-\frac{1}{2}\ln(4x-5)\) because it is a composition , the log of something for \(7x\) that is just a line with slope 7, derivative is 7

  35. monroe17
    • 2 years ago
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    woah im lost now.

  36. monroe17
    • 2 years ago
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    i did the chain rule for ln(4x-5) not 1/2*ln(4x-5)

  37. satellite73
    • 2 years ago
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    yea it is right what you wrote is correct

  38. satellite73
    • 2 years ago
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    the "minus one half" is just a constant, leave it there like you did

  39. monroe17
    • 2 years ago
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    okay, so where do i go from there. that fraction you did confused me.

  40. satellite73
    • 2 years ago
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    the only mistake in your answer was that you left \(7x\) there, when the derivative of \(7x\) is \(7\) everything else was right

  41. satellite73
    • 2 years ago
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    you wrote this 7x-1/2*(1/4x-5)*4

  42. monroe17
    • 2 years ago
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    oh i get it now!

  43. satellite73
    • 2 years ago
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    the 7x should be 7

  44. satellite73
    • 2 years ago
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    and the four is in the numerator, cancels with the 2 in the denominator

  45. monroe17
    • 2 years ago
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    derivative of 7x is 7. 1/2 of 4 is 2. 7-2/4x-5

  46. satellite73
    • 2 years ago
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    so it should look more like \(7-\frac{2}{4x-5}\)

  47. satellite73
    • 2 years ago
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    yup you got it

  48. monroe17
    • 2 years ago
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    thanks! im just a little slow sometimes lol

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