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I just need my work checked, derivatives

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    I just need to find the derivatives of the following:
    [a] y=ln(x^4 sin^2 x)
    f(x) = x^2 lnx, also find f'(1)
    [c] x^y = y^x

    2. Relevant equations
    See above


    3. The attempt at a solution
    [a] y' = [(4x^3)(sin^2x) + (x^4)(2sinxcosx)]/(x^4sin^2x) = (4sinx + 2xcosx)/(xsinx)

    f(x)' = (2x)(lnx) + (x^2)(1/x) = 2xlnx + x
    f'(1) = 2(1)ln(1) + 1 = 2(0) + 1 = 1

    [c] ln(x^y) = ln(y^x)
    ylnx = xlny
    d/dx ( ylnx) = d/dx (xlny)
    (1)(y')(lnx) + (y)(1/x) = (1)(lny) + (x)(1/y)(y')
    lnxy' + y/x = lny + xy'/y
    lnxy' - xy'/y = lny = y/x
    y'(lnx - x/y) = lny - y/x
    y' = (lny - y/x)/(lnx - x/y)
     
  2. jcsd
  3. Mar 19, 2007 #2
    Ridiculously boring. Who gives you those exercises?
     
  4. Mar 19, 2007 #3

    JasonRox

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    Professors who KNOW that their students are doing derivatives for the first time.
     
  5. Mar 19, 2007 #4
    LOL...are my answers correct?
     
  6. Mar 19, 2007 #5

    cristo

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    Good point; we all had to learn differentiation at some point. There's no need for such a pompous reply.
    Yes, presuming that the last question is asking for the derivative wrt x.
     
  7. Mar 19, 2007 #6
    Great, thank you!
     
  8. Mar 19, 2007 #7

    HallsofIvy

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    By the way, for [a] y=ln(x^4 sin^2 x)
    I would write y= 4 ln x+ 2 ln sin x and get
    [tex]y'= \frac{4}{x}+ 2\frac{cos x}{sin x}[/tex]
    Can you show that is the same as your answer?
     
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