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Logarithmic Differentiation

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    I've tried both and i) and ii) however something just doesn't seem right. Can somebody tell me if my method is correct?

    i) [​IMG] ii)[​IMG]
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2


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    Staff Emeritus
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    You forgot to differentiate.
  4. Apr 21, 2007 #3
    I've got a total mindblock :yuck: , can you explain a little further?
  5. Apr 21, 2007 #4
    differentiate both right and left hand side...
  6. Apr 21, 2007 #5
    Oh yeh! I tried by using chain rule (Let sinx^cosx = u)
    therefore i get :

    (1/y)*(dy/dx) = (1/u)*[d/dx(sinx^cosx)]

    How would I go about diff-ing sinx^cosx?
  7. Apr 21, 2007 #6
    consider cosx as u and log(sinx) as v, then diff by the product rule (udv+vdu)
    In diff. v, use the chain rule
  8. Apr 21, 2007 #7
    Ah! so much easier!
  9. Apr 21, 2007 #8
    Ok, after that tip from f(x) I got what I beleive is the final answer. Can this be simplified any further?

    Last edited: Apr 21, 2007
  10. Apr 21, 2007 #9

    D H

    Staff: Mentor

    Your work looks good up until very last step. Try that final step one more time.
  11. Apr 21, 2007 #10
    silly me! :rofl:

    2nd time lucky!
    Last edited: Apr 21, 2007
  12. Apr 21, 2007 #11

    D H

    Staff: Mentor

    Much better.
  13. Apr 21, 2007 #12
    I'm not too sure how to go about part (ii)
    So far i've converted the square roots to ^1/2 and expanded out to get

    I then let that equal u (chain rule) however the product rule within u seems to get pretty messy/complicated.

    Is this the correct method?
  14. Apr 21, 2007 #13


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    If [itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex] then let [itex]u=x^{3/2} [/itex]and [itex]v=(1-x^2)^{1/4} [/itex] so that y'=uv'+u'v, remembering that to calculate v' you will need to use the chain rule.
    Last edited: Apr 21, 2007
  15. Apr 22, 2007 #14
    Ok, so I now know how to diff [itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex] but shouldn't I take log of both sides before diff-ing? In that case I need to figure out the derivative of ln[[itex]y=x^{3/2}[(1-x^2)^{1/4}],[/itex]]
  16. Apr 23, 2007 #15
    Bump, can anybody help?
  17. Apr 23, 2007 #16


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    You could but you don't have to. Typically "logarithmic differentiation is used when you have a function of x in the exponent.

    Just use the product rule:
    [itex]y'= (x^{3/2})'(1- x^2)^{1/4}+ x^{3/2}[(1-x^2)^{1/4}]'[/itex]
    [itex]= (3/2)x^{1/2}(1- x^2)^{1/4}+ (x^{3/2})(1/4)(1-x^2)^{-3/4}[/itex]

    If you take the logarithm of both sides, you get [itex]ln y= (3/2)ln(x)+ (1/4)ln(1+ x^2)[/itex]. Now [itex](1/y)y'= 3/(2x)+ 2x/(4(1+x^2))[/itex]. That righthand side is much simpler but you still have to multiply by y.
  18. Apr 23, 2007 #17


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    Another way to differentiate the second expression is to change it to a single n-root, like this:
    [tex]\sqrt{x ^ 3 \sqrt{1 - x ^ 2}} = \sqrt{\sqrt{x ^ 6 (1 - x ^ 2)}} = \sqrt[4]{x ^ 6 - x ^ 8}[/tex]
    Let u = x6 - x 8. Now, the whole expression becomes [tex]\sqrt[4]{u}[/tex], you can apply the chain rule and finish the problem. Can you go from here? :)
  19. Apr 24, 2007 #18
    I think ill go with this method. So using chain rule I get dy/dx = (1/4)((x^6-x^8)^(-3/4))*(6x^5-8x^7)

    Hope you can make that out. So where does the logarithmic part come into it? Do I now take log of both sides?

    I'm quite lost:confused:
  20. Apr 24, 2007 #19


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    Nope, you can differentiate it normally as you've just done.

    Or you can tuse logarithmic differentiation as HOI has suggested, i.e, it goes like this:

    [tex]y = \sqrt{x ^ 3 \sqrt{1 - x ^ 2}}[/tex]

    Taking log of both sides yields:

    [tex]\ln y = \ln \left( \sqrt{x ^ 3 \sqrt{1 - x ^ 2}} \right) = \frac{1}{2} \ln \left( x ^ 3 \sqrt{1 - x ^ 2} \right) = \frac{1}{2} \left( \ln (x ^ 3) + \ln( \sqrt{1 - x ^ 2} ) \right)[/tex]

    Now, you can differentiate both sides to find y'.

    There are many ways to approach a differentiation problem, so just choose the one that you like best.
  21. Apr 25, 2007 #20
    That's very helpful! Had to navigate through the diff of [tex]\frac{1}{2} \left( \ln (x ^ 3) + \ln( \sqrt{1 - x ^ 2} ) \right)[/tex]. Was a bit tricky but I think I got it...

    Can you please confirm?
    Last edited: Apr 25, 2007
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