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Thread: Derivate

  1. #1
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    Talking Derivate

    Hello

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

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

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

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

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

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

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

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

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

    Let's move on...

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

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

    I think it looks correct, can it be simplified?

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

    Anything further?

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

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

    Thanks for all replies.
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  2. #2
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    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.

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

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

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

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

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

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

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

    $\displaystyle ln(k*x^n)=ln(k)+n ln(x)$

    So the derivate should be n/x


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

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

    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
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  3. #3
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    It could be easier to tackle this with the quotient rule;

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

    With a bit of re-arranging i got;

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

    Sorry my latex is poor.
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  4. #4
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    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.

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

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

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

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

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

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

    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...

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

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

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

    $\displaystyle -\frac{21}{x} - \frac {21}{x}$

    $\displaystyle -\frac{42}{x}$

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

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

    can $\displaystyle \frac{ln(x)}{x}$ be simplified somehow?
    Last edited by λιεҗąиđ€ŗ; Mar 9th 2008 at 07:22 AM.
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  5. #5
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    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.

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

    Quotient Rule:

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


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


    $\displaystyle = \frac{ -\ln 5}{3x \, (\ln x)^2}$.
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  6. #6
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    ahhh i see the mistake λιεҗąиđ€ŗ

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

    Goes to;

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

    NOT

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