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Math Help - derivative involving "ln"

  1. #1
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    derivative involving "ln"

    If f(x) = ln(ln(1-x)), then f'(x) =

    a) -1/ln(1-x)
    b) 1/[(1-x)ln(1-x)]
    c) 1/[(1-x)^2]
    d) -1/[(1-x)ln(1-x)]
    e) -1/[ln(1-x)^2]
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by DINOCALC09 View Post
    If f(x) = ln(ln(1-x)), then f'(x) =

    a) -1/ln(1-x)
    b) 1/[(1-x)ln(1-x)]
    c) 1/[(1-x)^2]
    d) -1/[(1-x)ln(1-x)]
    e) -1/[ln(1-x)^2]
    You should go over the chain rule and derivatives of logarithms.

    f'(x) = \frac{-1}{(1-x)ln(1-x)}
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    how does the double ln work [ln(ln(...]
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    Quote Originally Posted by DINOCALC09 View Post
    how does the double ln work [ln(ln(...]
    If x = ln(ln(y)), then x is basically the power you put on e^e (not e squared, but e^e) to get y.
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  5. #5
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    All 'ln's work the same. It's just an application of the chain rule.

    [f(g(h(x)))]' = f'(g(h(x)))*g'(h(x))*h'(x)

    If f, g, and h are all the same, it just looks funny, it doesn't work any differently.
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  6. #6
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    \ln \ln x ?

    you use the chain rule.

    u = \ln x
    \frac{du}{dx} = \frac{1}{x}

    \ln \ln x  \rightarrow  \ln u

     \frac {d}{dx} ( \ln u ) = \frac{1}{u} \cdot \frac{du}{dx}

     \frac {d}{dx} ( \ln u ) = \frac{1}{u} \cdot \frac{1}{x}
     \frac {d}{dx} ( \ln \ln x ) = \frac{1}{x \ln x}
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    \mbox{derivative of} \ln (\mbox{ANYTHING}) = \frac 1{\mbox{ANYTHING}} \times \mbox{derivative of ANYTHING}

    Just a matter of applying the chain rule as others said. and of course, we assume 1/(anything) is defined and what not...
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  8. #8
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    Quote Originally Posted by DINOCALC09 View Post
    If f(x) = ln(ln(1-x)), then f'(x) =

    a) -1/ln(1-x)
    b) 1/[(1-x)ln(1-x)]
    c) 1/[(1-x)^2]
    d) -1/[(1-x)ln(1-x)]
    e) -1/[ln(1-x)^2]
    Ya others already said it, but if you are new to this,solve any such problem like this :
    When you see ln (>something here<), at first step dont worry about >something here< just write 1/>something here< on paper. Now look at that >somethin here< it could be another function, say sin(>more something<), again forget what ever is inside parentheses just write derivative of sin (that is cos....) Keep on proceeding until you run out of functions.

    Example:
    say you are asked to differentiate  (x+\ln(\sin(x^2+1)^3)))^4

    Yuck! Ugly function but don't worry...

    Start Identifying functions from the >outermost< parts.
    What I do?
    ***I tell myself, hey looks like some Y^4 ,so I write down: 4(Y)^3(...........). Now let's fill "............"
    ***Now I see what's inside, hey looks like some x+ \ln Y(This is not the old Y, I am using Y as a variable!). I say to myself, I know it's derivative, it's 1 + \frac1{Y}, so I write down 4(x+\ln(\sin(x^2+1)^3)))^3 (1 + \frac{1}{\sin(x^2+1)^3}(............)).
    ***Proceeding like this I notice next up is sin Y and its derivative is cos Y. So now I can happily write 4(x+\ln(\sin(x^2+1)^3)))^3 (1 + \frac{1}{\sin(x^2+1)^3}(\cos(x^2+1)^3)(.........))

    Hope you get the idea
    I know it looks complicated but it is just a matter of practice. So can you complete my answer ??
    Hope you do well
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  9. #9
    GAMMA Mathematics
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    Good explanation of the chain rule guys! I always liked using the word "junk" when teaching the chain rule. Junk can contain a carrot, and you would have to take the derivative of carrot and multiply that by junk.
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  10. #10
    Lord of certain Rings
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    Quote Originally Posted by colby2152 View Post
    Good explanation of the chain rule guys! I always liked using the word "junk" when teaching the chain rule. Junk can contain a carrot, and you would have to take the derivative of carrot and multiply that by junk.


    In fact, I used to explain it with the same vegetable!!!. That's really a strange co-incidence OR it's a common mathematical parlance
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  11. #11
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    Quote Originally Posted by DINOCALC09 View Post
    If f(x) = ln(ln(1-x)), then f'(x) =

    a) -1/ln(1-x)
    b) 1/[(1-x)ln(1-x)]
    c) 1/[(1-x)^2]
    d) -1/[(1-x)ln(1-x)]
    e) -1/[ln(1-x)^2]
    Instead takin' the derivative quickly, do the make-up:

    f(x) = \ln (\ln (1 - x))\iff e^{f(x)}  = \ln (1 - x).

    Now you can easily contemplate its derivative

    f'(x)e^{f(x)}  =  - \frac{1}<br />
{{1 - x}}\,\therefore \,f'(x) =  - \frac{1}<br />
{{(1 - x)\ln (1 - x)}}.

    And we're done.
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