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Derivative of Exponential and Logarithmic Functions

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data

    PROBLEM 1
    Find slope of tangent line at (1,0)

    g(x) = x2lnx

    Solution:

    g'(x) = (x2lnx)'
    g'(x) = x2[1/x . (x)']
    g'(x) = x2 . 1/x
    g'(x) = x

    M tan/(1,0) = 0

    PROBLEM 2

    Use implicit differentiation:

    yex+ y2 = 7lny - x3

    Solution:

    (ex)(y') + 2y(y') - 7(y'/y) = -ye - 3x2

    y'(ex + 2y - 7/y) = -yex - 3x2
    y' = -yex - 3x2/(ex + 2y - 7/y)
     
    Last edited: Jan 20, 2013
  2. jcsd
  3. Jan 20, 2013 #2

    Ray Vickson

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    Your derivative of ##x^2 \ln(x)## is wrong.
     
  4. Jan 20, 2013 #3
    Thanks for the reply. That's why posted both problems. I want to understand why.
    If the derivative of 7lny on the second problem is 7[(1/y . (y)']
    then why x^2 ln(x) isnt = x^2[1/x . (x)'] ?
     
  5. Jan 20, 2013 #4
    OHHHH I THINK I GOT IT! Am I supposed to use the product rule?
     
  6. Jan 20, 2013 #5

    SammyS

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    The derivative of 7 is zero, because 7 is a constant . x2 is not a constant.

    Use the product rule.
     
  7. Jan 20, 2013 #6
    Ok, if so...this is how I'd do it:

    g(x) = x2lnx

    g'(x) = (x2)(1/x) + (lnx)(2x)
    g'(x) = x + 2xlnx
    g'(x) = 1 + 2(1)ln(1)
    g'(x) = x + 2ln1
    g'(x) = 1 + 0 =1
     
  8. Jan 20, 2013 #7
    Thanks a lot!
     
    Last edited: Jan 20, 2013
  9. Jan 20, 2013 #8
    I have another question:

    Is it wrong to solve the problem this way:

    y = ln (√x3- 5)

    y' = [(x3 - 5)1/2]'/[x3 - 5]

    y' = 1/2(x3 - 5)-1/2 (x3 - 5)'/ (x3 - 5)

    y' = 3x2/ 2[(√x3 - 5)(√x3 - 5)]

    y' = 3x2/ 2(x3 - 5)

    Instead of:

    y = ln (x3 - 5)1/2 = 1/2 ln (x3 - 5)
    y' = 1/2 . (x3 - 5)'/(x3 - 5)
    y' = 3x2/2(x3 - 5)

    Can I solve it either way or could the first method cause me further problems with other exercises?
     
    Last edited: Jan 20, 2013
  10. Jan 20, 2013 #9

    SammyS

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    It's best to start a new thread for a new problem.

    Also, use sufficient parentheses.
    √(x3-5) ≠ √x3-5 .​

    As for your question: Either method is fine, but writing ln(ur) as (r)ln(u) reduces the number of times you need to use the chain rule.
     
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