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Math Help - Natural log and derivates

  1. #1
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    Natural log and derivates

    I am trying to find dy/dx
    The problem states: e^y = 1 + (X^2)(y)..My calculus book doesn't really explain how to get the derivative in terms I am making any sense of.

    Can I rewrite this as?

    y= ln(1+(x^2)(y))

    What rule would I need to apply to get the next step? would this be a chain rule problem or is their some special log rule I am missing. Any help would be appreciated. Thanks
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  2. #2
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    I'd differentiate it implicitly rather than trying to re-arrange everything...that way it's a lot easier:

    Just differentiate everything on both sides with respect to x:
    <br />
\begin{array}{l}<br />
 e^y  = 1 + yx^2  \\ <br />
 e^y \frac{{dy}}{{dx}} = 2xy + x^2 \frac{{dy}}{{dx}} \\ <br />
 \end{array}<br />


    Upon re-arranging:
    <br />
\begin{array}{l}<br />
 \frac{{dy}}{{dx}}\left( {e^y  - x^2 } \right) = 2xy \\ <br />
 \frac{{dy}}{{dx}} = \frac{{2xy}}{{e^y  - x^2 }} \\ <br />
 \end{array}<br />
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by free_to_fly View Post
    I'd differentiate it implicitly rather than trying to re-arrange everything...that way it's a lot easier:

    Just differentiate everything on both sides with respect to y:
    <br />
\begin{array}{l}<br />
 e^y  = 1 + yx^2  \\ <br />
 e^y \frac{{dy}}{{dx}} = 2xy + x^2{\color{red}\frac{\,dy}{\,dx}}  \\ <br />
 \end{array}<br />

    Upon re-arranging:
    <br />
\frac{{dy}}{{dx}} = \frac{{2xy + x^2 }}{{e^y }}<br />
    You forgot a \frac{\,dy}{\,dx}!!

    The answer should be \frac{\,dy}{\,dx}=\frac{2xy}{e^y-x^2}

    --Chris
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  4. #4
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    Yep I realized that soon as I posted it...what a silly mistake lol. Forgot it was implicit on the RHS too.
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    You forgot a \frac{\,dy}{\,dx}!!

    The answer should be \frac{\,dy}{\,dx}=\frac{2xy}{e^y-x^2}

    --Chris

    Thank you both. I do have one more question though if you guys don't mind. How does the X^2 become in the denominator?
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  6. #6
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    Actually I just figured it out. haha..Im guessing you brought the dy/dx over and factored it out then divided. I don't think before asking questions sorry.
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