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Math Help - 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 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

    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-. <br />
y = \log _5 \left( {3x - 7} \right) = \frac{{\ln \left( {3x - 7} \right)}}<br />
{{\ln 5}}<br />

    Whithot to forget the chain rule:

    <br />
y' = \frac{1}<br />
{{\ln 5}} \cdot \frac{1}<br />
{{3x - 7}} \cdot 3<br />

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

    <br />
y = 9^{2t}  = e^{2t\ln 9}  \Rightarrow y' = 2\ln 9 \cdot e^{2\ln 9 \cdot t} <br />

    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.:

     \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:

    <br />
 { 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-. <br />
y = \log _5 \left( {3x - 7} \right) = \frac{{\ln \left( {3x - 7} \right)}}<br />
{{\ln 5}}<br />

    Whithot to forget the chain rule:

    <br />
y' = \frac{1}<br />
{{\ln 5}} \cdot \frac{1}<br />
{{3x - 7}} \cdot 3<br />



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

    <br />
y = 9^{2t}  = e^{2t\ln 9}  \Rightarrow y' = 2\ln 9 \cdot e^{2\ln 9 \cdot t} <br />

    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 e^{2\ln 9 \cdot t}<br />
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: <br />
y = 3x = x + x + x \Rightarrow y' = 1 + 1 + 1 = 3<br />
but is more easye leave outside the constant

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

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

    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: <br />
\int {\frac{1}<br />
{{\sqrt {1 - y^2 } }}dy} <br />
whic is easy to calculate

    Then you note:

    <br />
\begin{gathered}<br />
   - x^2  + 4x - 1 =  - x^2  + 4x - 1 + 4 - 4 =  - x^2  + 4x - 4 + 3 \hfill \\<br />
   =  - \left( {x^2  - 4x + 4} \right) + 3 = 3 - \left( {x - 2} \right)^2  = 3\left\{ {1 - \frac{{\left( {x - 2} \right)^2 }}<br />
{3}} \right\} \hfill \\ <br />
\end{gathered} <br />

    Then:

    <br />
\int {\frac{1}<br />
{{\sqrt { - x^2  + 4x - 1} }}} dx = \int {\frac{1}<br />
{{\sqrt {3\left\{ {1 - \frac{{\left( {x - 2} \right)^2 }}<br />
{3}} \right\}} }}} dx = \frac{1}<br />
{{\sqrt 3 }}\int {\frac{1}<br />
{{\sqrt {1 - \left( {\frac{{x - 2}}<br />
{{\sqrt 3 }}} \right)^2 } }}dx} <br />

    Now the substitution is in all books:

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

    restituting:

    <br />
\int {\frac{1}<br />
{{\sqrt { - x^2  + 4x - 1} }}} dx = \arcsin \left( {\frac{{x - 2}}<br />
{{\sqrt 3 }}} \right) + C<br />
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