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Thread: Derivatives and Integrals of Logarithms

  1. #1
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    Derivatives and Integrals of Logarithms

    I really need someone to atleast give me the answers to some of these problems because I have no clue as how to do half of them they're so much harder than the problems in the book and I have a test tomarrow
    1. Y=log5(3x-7)
    2. y=9^2t
    3. y=(x^2)(e^-2/x)
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  2. #2
    Member Nacho's Avatar
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    what you want? integrate those functions? o find your derivatives?

    P.D: the last function is $\displaystyle y=x^2 \cdot e^{\frac{-2}{x}}$ ?
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  3. #3
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    Sorry about not being clear but for those I need the derivatives and the last one is

    $\displaystyle e^(-2/x)$

    If you cant tell the -2/x is the power
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  4. #4
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    Double post sorry
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  5. #5
    Member Nacho's Avatar
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    do you know base change in logarithms?

    1-. $\displaystyle
    y = \log _5 \left( {3x - 7} \right) = \frac{{\ln \left( {3x - 7} \right)}}
    {{\ln 5}}
    $

    Whithot to forget the chain rule:

    $\displaystyle
    y' = \frac{1}
    {{\ln 5}} \cdot \frac{1}
    {{3x - 7}} \cdot 3
    $

    2-. You must know the identity: $\displaystyle a=e^{\ln a}$

    $\displaystyle
    y = 9^{2t} = e^{2t\ln 9} \Rightarrow y' = 2\ln 9 \cdot e^{2\ln 9 \cdot t}
    $

    Can you do the 3?
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  6. #6
    Member Mentia's Avatar
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    Im sure you know that the derivative of the natural log of x is 1/x, i.e.:

    $\displaystyle \frac{ d}{dx } ln(x) = \frac{1 }{x } $,

    so you have a base 5 log in your first question. You should convert that to log base e so that you can do the derivative:

    $\displaystyle
    { log}_{5 } (x) = \frac{ ln(x)}{ln(5) }$

    You can get it from here I'm sure.
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  7. #7
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    Quote Originally Posted by Nacho View Post
    do you know base change in logarithms?

    1-. $\displaystyle
    y = \log _5 \left( {3x - 7} \right) = \frac{{\ln \left( {3x - 7} \right)}}
    {{\ln 5}}
    $

    Whithot to forget the chain rule:

    $\displaystyle
    y' = \frac{1}
    {{\ln 5}} \cdot \frac{1}
    {{3x - 7}} \cdot 3
    $



    2-. You must know the identity: $\displaystyle a=e^{\ln a}$

    $\displaystyle
    y = 9^{2t} = e^{2t\ln 9} \Rightarrow y' = 2\ln 9 \cdot e^{2\ln 9 \cdot t}
    $

    Can you do the 3?
    Whats the answer to the first one? Shouldnt the Ln5 go away since it's a constant?

    For the second one does the $\displaystyle e^{2\ln 9 \cdot t}
    $ need to be in the answer?

    I have alot of questions that I need help with by the way im really struggling in this class and praying to pass with a C.
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  8. #8
    Member Nacho's Avatar
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    I try explain you, but my english is a little basic ok?

    In you first question, the constant, multiplying the variable, keeps

    for example: $\displaystyle
    y = 3x = x + x + x \Rightarrow y' = 1 + 1 + 1 = 3
    $ but is more easye leave outside the constant

    but is diferent to derivate a constant! which always worths zero$\displaystyle
    y = 3x + 4 \Rightarrow y' = 3 \cdot 1 + 0
    $

    To second question: yes, you note $\displaystyle 2 \ln 9$ is a constan, then is same to derivate $\displaystyle y=e^c$ your derivative is $\displaystyle c \cdot e^c$

    any question, you ask again
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  9. #9
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    Ok here's another one its an integral

    $\displaystyle dx/(-x^2+4x-1)^(1/2)$

    The the (-x^2 +4X-1) is under a square root. Dont worry your English is fine and thanks for helping me out.
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  10. #10
    Member Nacho's Avatar
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    When I see this integral, my intuition say that is an integral of the form: $\displaystyle
    \int {\frac{1}
    {{\sqrt {1 - y^2 } }}dy}
    $ whic is easy to calculate

    Then you note:

    $\displaystyle
    \begin{gathered}
    - x^2 + 4x - 1 = - x^2 + 4x - 1 + 4 - 4 = - x^2 + 4x - 4 + 3 \hfill \\
    = - \left( {x^2 - 4x + 4} \right) + 3 = 3 - \left( {x - 2} \right)^2 = 3\left\{ {1 - \frac{{\left( {x - 2} \right)^2 }}
    {3}} \right\} \hfill \\
    \end{gathered}
    $

    Then:

    $\displaystyle
    \int {\frac{1}
    {{\sqrt { - x^2 + 4x - 1} }}} dx = \int {\frac{1}
    {{\sqrt {3\left\{ {1 - \frac{{\left( {x - 2} \right)^2 }}
    {3}} \right\}} }}} dx = \frac{1}
    {{\sqrt 3 }}\int {\frac{1}
    {{\sqrt {1 - \left( {\frac{{x - 2}}
    {{\sqrt 3 }}} \right)^2 } }}dx}
    $

    Now the substitution is in all books:

    $\displaystyle
    \begin{gathered}
    \sin z = \frac{{x - 2}}
    {{\sqrt 3 }} \Rightarrow \cos zdz = \frac{1}
    {{\sqrt 3 }}dx \hfill \\
    \hfill \\
    \frac{1}
    {{\sqrt 3 }}\int {\frac{1}
    {{\sqrt {1 - \sin ^2 z} }} \cdot \sqrt 3 \cos z} dz = \int {\frac{{\cos z}}
    {{\cos z}}dz} = z + C \hfill \\
    \end{gathered}
    $

    restituting:

    $\displaystyle
    \int {\frac{1}
    {{\sqrt { - x^2 + 4x - 1} }}} dx = \arcsin \left( {\frac{{x - 2}}
    {{\sqrt 3 }}} \right) + C
    $
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