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Limits, Derivatives, and Trig

  1. Oct 19, 2007 #1
    I'm not that good with trig problems, I don't know what it is. Here are some answers I got, just wondering if they are right.

    1) If x cos (y) + y cos (x) = 1, find an expression for dy/dx

    Ok so using the product rule I got
    (x)(-sin y)+(1)(cos y) + (y)(-sin x)+(1)(cos x) = 0 =>
    -x sin (y) + cos (y) - y sin (x) + cos (x)

    2) f(t) = tan(sin t²)

    f'(t) = sec² (sin t²)(2sin t)(cos t)
    can this be reduced? do I have the brackets right?

    3) Find the value of lim x->0 (tan 2x)/x

    I plugged sin/cos in for tan and got
    ( sin 2x/cos 2x ) / (x) =>
    ( sin 2x/cos 2x ) * (1/x)

    but now I'm stuck

    4) Find y' if y = log (base 3) (x²e^x)
    y' = (x²e^x + 2xe^x) / (x²e^x)(ln 3)
    factor out x²e^x
    y' = (x+2)/(x ln 3)
     
  2. jcsd
  3. Oct 19, 2007 #2

    mjsd

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    is x or y a function of each other?

    your f'(t) is wrong

    use L'Hopital's rule (ans: 2)

    wrong again

    EDIT: sorry this one is ok (ignore my previous comment)
     
    Last edited: Oct 19, 2007
  4. Oct 19, 2007 #3
    1) Working...

    2) How far off am I? All you do is apply the chain rule right?

    3) Is there a way to solve this with out l'hospital's rule? I don't think we've learned it yet...
     
  5. Oct 19, 2007 #4
    1) Since y is a function of x
    -xy'sin(y) + cos(y) - ysin(x) + y'cos(x) = 0
    y'(-xsin(y)) + y'cos (x) = ysin(x) - cos(y)
    y' = (ysin (x) - cos (y)) / (-xsin(y) + cos(x))
     
  6. Oct 19, 2007 #5

    mjsd

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    yes..you were close

    use implicit differentiation... and you have got it worked out.
     
  7. Oct 19, 2007 #6

    mjsd

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    for 3)
    note that
    [tex]\tan(z) = z + \frac{z^3}{3} + \frac{2z^5}{15} +\frac{17 z^7}{315} +O(z^8)[/tex]
    divide thru and then sub in value....
     
  8. Oct 19, 2007 #7
    2) f'(t) = sec²(sin t²)(cos t²)(2t)
     
  9. Oct 19, 2007 #8

    mjsd

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    yes.
     
  10. Oct 19, 2007 #9
    Thanks for your help

    For 3) I ended up getting

    lim x-> 0 (sin x /x)+(sin x /x) / cos 2x = 2
     
  11. Oct 19, 2007 #10

    mjsd

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    here you have used sin x/x ->1 as x->0 which is usually proved using L'Hopital rule....or of course can be seen by expanding into power series
     
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