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    Ive got a question that I just can't seem to get the answer to:
    find dy/dx of  \sqrt {xy} + \sin x + \cos y = 0

    The final answer is  \frac{y + 2 \sqrt { \cos x }} {2 \sqrt {xy}  \sin y - x} but I am not entirely sure on how to get to this stage.

    My attempt:
    I think I worked the derivative of
     \sqrt {xy} to be  \frac{1}{2 \sqrt {xy}} * (y + x * \frac{dy}{dx})
    The derivative of
     \sin x to be  cosx (obviously)
    And finally
     \cos y to be  (- \sin y \frac{dy}{dx})
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    Move the sin(x) term to the RHS (it`s easier to deal then with factorising the y ' (x) terms)

    differentiate the entire thing, so you get:

    \displaystyle  \frac{y^{1/2}}{2x^{1/2}}+\frac{x^{1/2}}{2y^{1/2}}y'- \sin(y) y ' = -\cos(x)

    then multiply the entire equation by:

    2x^{1/2}y^{1/2}

    and see where that gets you..

    (ooh, and factor out the y' on the LHS, moving the 1st term in the 1st line above to the RHS)

    (EDITED)
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    I've tried it again, using methods that you showed and have ended up with (-y-2cos(x)(xy)^1/2) / (-2((xy)^1/2) sin(y)-x)
    Apologies, LaTex isn't working for some reason.
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    your answer is correct apart from a minus sign infront of 2root(x)root(y) in the denominator (maybe a small mistake somewhere) - just take a factor of -1 out from the numerator in the answer you gave, THEIR numerator is incorrect as it doesn`t have the root(xy) term (the root(cos(x) term is wrong also)
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    That's how I'll deal with all problems in the future. If my answer doesn't match theirs, their answer is wrong! :P But seriously, I think you're right, I see no possible way of getting root cos anywhere in that equation. Thanks for the help!
 
 
 
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Updated: January 28, 2015
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