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Math Help - implicit differentiation con't ...01

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

    Problem:

    4cosxsiny = 1

    Solution attempt:


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

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

    simplify;
    -4sinxsiny+4cosxcosyy' = 0

    add 4sinxsiny to both sides;
    4cosxcosyy' = 4sinxsiny

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

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

    Answer has been verified, thank you all.
    Last edited by Foxlion; March 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:

    4cosxsiny = 1

    Solution attempt:


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

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

    simplify and factor;
    -4sinxsiny+y'(4cos^2y) = 0
    I believe there is an error in this step, namely re-writing 4 \cos x \cos y as 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  cos^2x I'm such a fool.

    original post edited for corrections again thank you all.
    Last edited by Foxlion; March 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:

    4cosxsiny = 1

    Solution attempt:


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

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

    simplify;
    -4sinxsiny+4cosxcosyy' = 0

    add 4sinxsiny to both sides;
    4cosxcosyy' = 4sinxsiny

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

    apply trig identity  \frac{sinx}{cosx} = tanx and simplify;
     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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