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Thread: Derative

  1. #1
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    Derative

    How to find a derative by definition for such a function:

    $\displaystyle y=xln^2(2x+1)$
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  2. #2
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    Quote Originally Posted by OnHold View Post
    How to find a derative by definition for such a function:

    $\displaystyle y=xln^2(2x+1)$
    Have you made any attempt at this, using the product and the chain rule?
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  3. #3
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    Yes, I have. I need to find derative in the point (3)

    $\displaystyle ln^2(2h+2x+1)-(3/h)*ln^2(2x+1)$
    $\displaystyle ln(2h+7)-ln(7)^(3/h)(ln(2h+7)+ln7^(3/h)$
    There my knowledge finishes.
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  4. #4
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    Quote Originally Posted by OnHold View Post
    How to find a derative by definition for such a function:

    $\displaystyle y=xln^2(2x+1)$
    You need to use the product rule and the chain rule...

    Chain rule: $\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$.

    Product rule: $\displaystyle \frac{d}{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx}$.


    So in your case:

    $\displaystyle y = x[\ln{(2x + 1)}]^2$.

    Can you see that this is a product of functions? So you need to use the product rule with $\displaystyle u = x$ and $\displaystyle v = [\ln{(2x + 1)}]^2$.


    So $\displaystyle \frac{dy}{dx} = x\,\frac{d}{dx}\{[\ln{(2x + 1)}]^2\} + [\ln{(2x + 1)}]^2\,\frac{d}{dx}(x)$

    $\displaystyle = x\,\frac{d}{dx}\{[\ln{(2x + 1)}]^2\} + [\ln{(2x + 1)}]^2$.


    Now to find $\displaystyle \frac{d}{dx}\{[\ln{(2x + 1)}]^2\}$ you need to use the chain rule twice...

    Can you go from here?
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