Results 1 to 6 of 6

Math Help - Derivate

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    35

    Talking Derivate

    Hello

    I'd like some help on derive a function and simplifying the expression as much as possible.

    f(x) = \frac{ln(5x^7)}{ln(x^3)}
    ----

    Here's my idea on how it ends up...

    first, we can work a little bit with the function:

    f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}

    Ok this looks a bit easier, we'll get the derivate of an product.

    D(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)

    The general derivate of ln:
    D ln(kx^n) = \frac{1}{x^n}

    and \frac{1}{x} simply ends up as -\frac{1}{x^2}

    Let's move on...

    Let's derivate f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}

    \frac{1}{5x^7} \cdot 5 \cdot 7x^6 \cdot \frac{1}{ln(x^3)} + ln(5x^7) \cdot (- \frac{1}{ln(x^3)^2}) \cdot 3x^2

    I think it looks correct, can it be simplified?

    \frac {7}{x^7} \cdot x^4 \cdot \frac {1}{ln(x^3)} + (- \frac {ln(5x^7)}{ln(x^3)^2}) \cdot 3 |/x^2

    Anything further?

    \frac{7x^4}{x^7} - \frac {3ln(5x^7)}{ln(x^3)}  |\cdot ln(x^3)

    \frac{7}{x^3} - \frac{3ln(5x^7)}{ln(x^3)}

    Thanks for all replies.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hi


    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Hello

    I'd like some help on derive a function and simplifying the expression as much as possible.

    f(x) = \frac{ln(5x^7)}{ln(x^3)}
    ----

    Here's my idea on how it ends up...

    first, we can work a little bit with the function:

    f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}

    Ok this looks a bit easier, we'll get the derivate of an product.

    D(u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
    Until here it's ok

    The general derivate of ln:
    D ln(kx^n) = \frac{1}{x^n}
    I don't agree.

    ln(k*x^n)=ln(k)+n ln(x)

    So the derivate should be n/x


    Let's derivate f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}

    \frac{1}{5x^7} \cdot 5 \cdot 7x^6 \cdot \frac{1}{ln(x^3)} + ln(5x^7) \cdot (- \frac{1}{ln(x^3)^2}) \cdot 3x^2

    I think it looks correct, can it be simplified?
    It doesn't seem correct to me...

    Let's see the derivate of 1/(ln(x^3)). This is a function similar to 1/u(x). But the derivate of such a function is -u'(x)/uČ(x).
    Here, u(x)=ln(x^3)
    And the derivate of u(x) is not 3xČ but -3/x.

    Here is the mistake, try to write it down again
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2007
    Posts
    144
    It could be easier to tackle this with the quotient rule;

    \frac{v\frac{du}{dx}~-~u\frac{dv}{dx}}{v^2}

    With a bit of re-arranging i got;

    \frac{-3\ln5}{x(\ln(x^3))^2}

    Sorry my latex is poor.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2007
    Posts
    35
    Quote Originally Posted by Sean12345
    It could be easier to tackle this with the quotient rule;
    It might, but let's see what we get when we work a bit with the logarithms as Moo suggested.

    f(x) = \frac{ln(5x^7)}{ln(x^3)}

    f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}

    ln(5) + 7ln(x) \cdot \frac {1} {lnx^3}

    ln(5) + 7ln(x) \cdot (ln(1) - 3ln(x)) Error

    ln(5) + 7ln(x) \cdot (0 - 3ln(x))

    ln(5) + 7ln(x) \cdot - 3ln(x)

    The above seems reasonable, I've kept it to the basics and I would be suprised if I made an error.

    Let's derive that...

    \frac{7}{x} \cdot -3ln(x) + 7ln(x) \cdot - \frac{3}{x}

    can we divide by ln(x)? I think we can.

    \frac{7}{x} \cdot -3 + 7 \cdot - \frac{3}{x}

    -\frac{21}{x} - \frac {21}{x}

    -\frac{42}{x}

    Hmm not really sure if we can remove ln(x) just like that ... in that case

    -\frac{42ln(x)}{x}

    can \frac{ln(x)}{x} be simplified somehow?
    Last edited by λιεҗąиđ€ŗ; March 9th 2008 at 08:22 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Hello

    I'd like some help on derive a function and simplifying the expression as much as possible.

    f(x) = \frac{ln(5x^7)}{ln(x^3)}
    [snip]
    f(x) = \frac{\ln 5 + 7 \ln x}{3 \ln x}.

    Quotient Rule:

    f'(x) = \frac{(7/x) \, (3 \ln x) - (3/x) \, (\ln 5 + 7 \ln x)}{(3 \ln x)^2}


    = \frac{21 \ln x - 3 \ln 5 - 21 \ln x}{9x \, (\ln x)^2}


    = \frac{ -\ln 5}{3x \, (\ln x)^2}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    144
    ahhh i see the mistake λιεҗąиđ€ŗ

    f(x) = ln(5x^7) \cdot \frac{1}{ln(x^3)}

    Goes to;

    ln(5) \cdot \frac {1} {lnx^3} + 7ln(x) \cdot \frac {1} {lnx^3}

    NOT

    ln(5) + 7ln(x) \cdot \frac {1} {lnx^3}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivate
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2011, 07:30 AM
  2. Derivate
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 29th 2010, 05:55 PM
  3. [SOLVED] How to derivate these?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 16th 2010, 02:56 PM
  4. How to derivate this one?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 9th 2010, 07:00 PM
  5. Derivate.
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 6th 2006, 01:08 AM

Search Tags


/mathhelpforum @mathhelpforum