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amy0799

  • one year ago

if g(x)=x^2f'(x)

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  1. amy0799
    • one year ago
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    where \[f(x)=\frac{ x-cosx }{ x }\] find the average rate of change of g on [pi, 2pi]

  2. SolomonZelman
    • one year ago
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    differentiate the f(x).

  3. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(x) =\frac{x-\cos x}{x} }\) \(\large\color{black}{ \displaystyle f(x) =1-\frac{\cos x}{x} }\) then derivative of 1 is 0, and derivative of the second peace will go by quotient rule (or you can use product or logarithmic)

  4. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }\)

  5. SolomonZelman
    • one year ago
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    continue please.

  6. amy0799
    • one year ago
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    why did you take the 2nd derivative?

  7. SolomonZelman
    • one year ago
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    I didn't take the second derivative.

  8. amy0799
    • one year ago
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    oh wait I misread ur sentence, sorry

  9. SolomonZelman
    • one year ago
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    My function is: \(\large\color{black}{ \displaystyle f(x) =\frac{x-\cos x}{x} }\) then I simplify it algebraically, to get: \(\large\color{black}{ \displaystyle f(x) =1-\frac{\cos x}{x} }\) And now, We have to differentiate it (once). \(\large\color{black}{ \displaystyle f'(x) =0-\frac{(\cos x)'x-\cos x(x)'}{x^2} }\)

  10. SolomonZelman
    • one year ago
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    if it is not making sense, then you are always welcome to express your concerns regarding the problem.

  11. amy0799
    • one year ago
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    so then \[g(x)=x ^{2}(\frac{ xsinx+cosx}{ x ^{2} })\]

  12. amy0799
    • one year ago
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    @SolomonZelman

  13. SolomonZelman
    • one year ago
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    yes, very good, and you can tell that x² will can cancel

  14. SolomonZelman
    • one year ago
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    Then the remaining part is an algebraic task of fining the slope (of the secant) between \((\) 2π,f(2π) \()\) and \((\) π,f(π) \()\)

  15. amy0799
    • one year ago
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    I put 2pi and pi into xsinx+cosx?

  16. amy0799
    • one year ago
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    @SolomonZelman

  17. SolomonZelman
    • one year ago
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    Yes, that is what you do:)

  18. amy0799
    • one year ago
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    (2pi,1) (pi,-1) is that correct?

  19. SolomonZelman
    • one year ago
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    Yes

  20. SolomonZelman
    • one year ago
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    then the slope formula...

  21. amy0799
    • one year ago
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    pi/2 is the answer?

  22. SolomonZelman
    • one year ago
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    |dw:1442437619280:dw|

  23. SolomonZelman
    • one year ago
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    Not the other way around

  24. amy0799
    • one year ago
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    oh oops so 2/pi?

  25. SolomonZelman
    • one year ago
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    Yup

  26. amy0799
    • one year ago
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    do u mind helping me with another problem?

  27. SolomonZelman
    • one year ago
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    Yes, if I will be able to.

  28. amy0799
    • one year ago
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    find \[\lim_{h \rightarrow 0} \frac{ \csc(\frac{ \pi }{ 6 }+h)-\csc\frac{ \pi }{ 6 }}{ h } \] @SolomonZelman

  29. SolomonZelman
    • one year ago
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    Oh boy, first principles of differentiation.... (you will get to the point, when this will be only useful to perhaps prove derivtive of e^x, but I guess right now we can't apply the easy properties right... will go with that limit definition... i have to refresh my page. sorry)

  30. amy0799
    • one year ago
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    when I did this I got 2 and I'm not sure if that's correct

  31. SolomonZelman
    • one year ago
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    Something was going on with my PC, it is kind of off.

  32. SolomonZelman
    • one year ago
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    Ok, I think I am back.

  33. SolomonZelman
    • one year ago
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    With \(f(x)=\csc(x)\) \(\large\color{slate}{\displaystyle f'(x)=\lim_{h \rightarrow ~0}\frac{\csc(x+h)-\csc(x)}{h}}\) \(\large\color{slate}{\displaystyle f'(\pi/6)=\lim_{h \rightarrow ~0}\frac{\csc(\pi/6+h)-\csc(\pi/6)}{h}}\)

  34. SolomonZelman
    • one year ago
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    You don't get 2 for that. I really don't think so.

  35. SolomonZelman
    • one year ago
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    I get -2√3

  36. amy0799
    • one year ago
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    how?

  37. SolomonZelman
    • one year ago
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    I applied the chain rule: \(\displaystyle \frac{d}{dx} (\csc x)=\frac{d}{dx} \left((\sin x)^{-1}\right)=(-1)\sin^{-2}x \times \color{blue}{\cos x} \\=-\csc^2x \cos x=-\cot x\csc x\)

  38. SolomonZelman
    • one year ago
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    Then I plugged π/6 instead of x.

  39. SolomonZelman
    • one year ago
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    This is just for check...

  40. SolomonZelman
    • one year ago
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    But, I guess you have to use that lim h-> 0 rule, right?

  41. amy0799
    • one year ago
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    where did you get the cosx from?

  42. SolomonZelman
    • one year ago
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    have you studied the "Chain Rule" ?

  43. SolomonZelman
    • one year ago
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    If not, then don't worry about my application of the chain rule...

  44. amy0799
    • one year ago
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    yes

  45. SolomonZelman
    • one year ago
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    ok, so you want to do the limit?

  46. amy0799
    • one year ago
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    oh wait I see how you got it.

  47. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(\frac{\pi}{6}+h)-\csc(\frac{\pi}{6})}{h}}\)

  48. SolomonZelman
    • one year ago
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    Still want to do the limit?

  49. amy0799
    • one year ago
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    sure. I don't understand how you got \[-\frac{ 2 }{ \sqrt{3} }\]

  50. SolomonZelman
    • one year ago
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    I didn't.

  51. SolomonZelman
    • one year ago
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    I said tha I got -2√3

  52. SolomonZelman
    • one year ago
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    We have to do the limit, don't we?

  53. amy0799
    • one year ago
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    I believe so

  54. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\csc(x+h)-\csc(x)}{h}}\) \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{1}{\sin(x+h)}-\dfrac{1}{\sin(x)}}{h}}\) \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)-\sin(x+h) }{\sin(x+h)\sin(x)}}{h}}\) \(\large\color{slate}{\displaystyle \lim_{h \rightarrow ~0}\frac{\dfrac{\sin(x)-\sin(x)\cos(h) -\sin(h)\cos(x)}{[\sin(x)\cos(h) +\sin(h)\cos(x)]\sin(x)}}{h}}\)

  55. SolomonZelman
    • one year ago
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    this is too much for me... going through all of this limit.

  56. amy0799
    • one year ago
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    I'm sorta getting lost on what ur doing. does finding the limit get u \[-2\sqrt{3}?\]

  57. SolomonZelman
    • one year ago
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    |dw:1442439909186:dw|

  58. SolomonZelman
    • one year ago
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    If, finding the derivative through the chain rule, and plugging π/6 gives -2√3, then the limit must give the same.

  59. SolomonZelman
    • one year ago
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    I drew two identities that you will need to get rid of h in the denominator.... It is just not right tho, if I will type all of this for you, according to the site's policy-;9

  60. amy0799
    • one year ago
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    oh then I guess we don't have to find the limit. can you explain how you −2√3 from the chain rule?

  61. amy0799
    • one year ago
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    how did you get −cotxcscx from csc^2xcosx?

  62. SolomonZelman
    • one year ago
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    Do you get the essence of the chain rule? (before we go on to our problem)

  63. amy0799
    • one year ago
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    yes

  64. SolomonZelman
    • one year ago
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    Ok, so we are differentiating \(\csc x\) (with respect to x) we know that: \(\csc x = 1/\sin x = \left( \sin x \right)^{-1} \) right?

  65. amy0799
    • one year ago
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    right

  66. SolomonZelman
    • one year ago
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    So, we are going to differentiate *THAT* now.

  67. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] }\)

  68. SolomonZelman
    • one year ago
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    Just as, \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[ x ^{-1}\right]=(-1)\cdot x^{-1-1}=-x^{-2} }\) So is, \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1}=-\left( \sin x \right)^{-2}}\) However, there is a *BUT* to that.

  69. SolomonZelman
    • one year ago
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    We must multiply times the derivative of the inner function, times the derivative of sin x.

  70. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}\)

  71. SolomonZelman
    • one year ago
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    this in blue, is the CHAIN RULE, that we must apply

  72. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =(-1)\cdot \left( \sin x \right)^{-1-1} \color{blue}{\times \frac{d }{dx}\left[\sin x\right]}}\) \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \left( \sin x \right)^{-2} \color{blue}{\times \cos x}}\)

  73. SolomonZelman
    • one year ago
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    so far so good?

  74. amy0799
    • one year ago
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    yup

  75. SolomonZelman
    • one year ago
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    then, you just simplify it: \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \left( \sin x \right)^{-2} \color{blue}{\times \cos x}}\) \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\left( \sin x \right)^{-1}\right] =- \csc^2 x \color{blue}{\times \cos x}}\) and thus we know that: \(\large\color{black}{ \displaystyle \frac{d }{dx} \left[\csc x\right] =- \csc x \cot x}\)

  76. SolomonZelman
    • one year ago
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    then, you just plug in π/6, which should not be a problem.

  77. amy0799
    • one year ago
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    I don't understand how you got −csc^2x×cosx

  78. SolomonZelman
    • one year ago
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    well, sin\(^{-2}\)x is same as 1/sin²x, which is csc²x. And the one in front came from the power rule from the very beginning

  79. SolomonZelman
    • one year ago
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    And cos(x), as I explained, is the chain rule.

  80. amy0799
    • one year ago
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    oh ok. so I have 2 more problems I need help with, can u help me? if not that's ok.

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