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Thread: Problem with derivative y'=(lnˇ2 * lnˇ3 * lnˇ4 x)'= ?

  1. #1
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    Problem with derivative[Edited]

    Hello,
    I have a problem with derivative.Sorry,my topic is abit off :P,not use to doing this stuff at the computer.
    $\displaystyle y = ln^2ln^3 ln^4 x$
    }][[]]}
    The book says that the right answer is $\displaystyle y'=\frac{24ln ln^3 ln^4 x}{x(ln ln^4 x)lnx}$

    Thank you
    Last edited by Tafka; Oct 6th 2010 at 01:10 PM.
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  2. #2
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    Quote Originally Posted by Tafka View Post
    Hello,
    I have a problem with derivative.Sorry,my topic is abit off :P,not use to doing this stuff at the computer.
    $\displaystyle y = ln^2ln^3 ln^4 x$
    }][[]]}
    The book says that the right answer is $\displaystyle y'=\frac{24ln ln^3 ln^4 x}{x(ln ln^4 x)lnx}$
    This question involves multiple use of the chain rule. To make it manageable, split it into smaller sections. Start by putting $\displaystyle u = \ln^3(\ln^4x)$. Then $\displaystyle y = \ln^2u$. By the chain rule, $\displaystyle \frac{dy}{dx} = 2\ln u * \frac1u * \frac{du}{dx} = {\displaystyle\frac{2\ln\bigl(\ln^3(\ln^4x)\bigr)} {\ln^3(\ln^4x)}}\frac{du}{dx}.$

    Now you have to find $\displaystyle \frac{du}{dx}$. Do that in the same way, putting $\displaystyle u = \ln^3v$, where $\displaystyle v = \ln^4x$, and using the chain rule to find $\displaystyle \frac{du}{dx}$ in terms of $\displaystyle \frac{dv}{dx}$. Then use the chain rule once more to find $\displaystyle \frac{dv}{dx}$. Finally, put everything together into a single expression and simplify it.

    BTW this is a classic example of a case where using parentheses correctly makes things much easier to read: $\displaystyle y = \ln^2\ln^3 \ln^4 x$ looks like a jumble, but $\displaystyle y = \ln^2\bigl(\ln^3 (\ln^4 x)\bigr)$ is a lot clearer.
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