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Thread: implicit differentiation con't ...01

  1. #1
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    implicit differentiation con't ...01

    Problem:

    $\displaystyle 4cosxsiny = 1$

    Solution attempt:


    create differentiation schema;
    $\displaystyle siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

    apply differentiation schema;
    $\displaystyle siny[-4sinx]+4cosx[cosyy'] = 0$

    simplify;
    $\displaystyle -4sinxsiny+4cosxcosyy' = 0$

    add $\displaystyle 4sinxsiny$ to both sides;
    $\displaystyle 4cosxcosyy' = 4sinxsiny$

    divide both sides by $\displaystyle 4cosxcosyy'$;
    $\displaystyle y' = \frac{4sinxsiny}{4cosxcosy}$

    apply trig identity $\displaystyle \frac{sinx}{cosx} = tanx$ and simplify;
    $\displaystyle y' = tanxtany$

    Answer has been verified, thank you all.
    Last edited by Foxlion; Mar 10th 2011 at 08:43 PM. Reason: Corrections
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  2. #2
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    Quote Originally Posted by Foxlion View Post
    Problem:

    $\displaystyle 4cosxsiny = 1$

    Solution attempt:


    create differentiation schema;
    $\displaystyle siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

    apply differentiation schema;
    $\displaystyle siny[-sinx]+4cosx[cosxy'] = 0$
    There are two problems with this line. You dropped a factor of 4 (that's just a typo, I think.) The other is the inclusion of an x instead of a y in the second term. It should read
    $\displaystyle 4siny[-sinx]+4cosx[cosyy'] = 0$

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    It should read
    $\displaystyle 4siny[-sinx]+4cosx[cosyy'] = 0$

    -Dan
    Done and done though I am still lost on steps 7 and 8. Is it necessary to expand with the trig idents?
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  4. #4
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    Quote Originally Posted by Foxlion View Post
    apply differentiation schema;
    $\displaystyle siny[-4sinx]+4cosx[cosyy'] = 0$

    simplify and factor;
    $\displaystyle -4sinxsiny+y'(4cos^2y) = 0$
    I believe there is an error in this step, namely re-writing $\displaystyle 4 \cos x \cos y$ as $\displaystyle 4 \cos ^2 y$.
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  5. #5
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    ah yes as dan's correction eliminates the existential possibility of $\displaystyle cos^2x$ I'm such a fool.

    original post edited for corrections again thank you all.
    Last edited by Foxlion; Mar 10th 2011 at 08:46 PM.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Foxlion View Post
    Problem:

    $\displaystyle 4cosxsiny = 1$

    Solution attempt:


    create differentiation schema;
    $\displaystyle siny[\frac{d}{dx}4(cosx)+4\frac{d}{dx}(cosx)] + 4cosx[\frac{d}{dx}siny]=\frac{d}{dx}1$

    apply differentiation schema;
    $\displaystyle siny[-4sinx]+4cosx[cosyy'] = 0$

    simplify;
    $\displaystyle -4sinxsiny+4cosxcosyy' = 0$

    add $\displaystyle 4sinxsiny$ to both sides;
    $\displaystyle 4cosxcosyy' = 4sinxsiny$

    divide both sides by $\displaystyle 4cosxcosyy'$;
    $\displaystyle y' = \frac{4sinxsiny}{4cosxcosy}$

    apply trig identity $\displaystyle \frac{sinx}{cosx} = tanx$ and simplify;
    $\displaystyle y' = tanxtany$

    Answer has been verified, thank you all.
    Please do not alter the original question. (See Forum Rule #7 here.) We all do some amount of it, but the amount of editing you have done here makes it hard to see what you did and why all these posts exist. It can get very confusing.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Please do not alter the original question. We all do some amount of it, but the amount of editing you have done here makes it hard to see what you did and why all these posts exist. It can get very confusing.

    -Dan
    It can, I apologize. From now on I shall leave the original intact save for purely typographical corrections.
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