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Math Help - basic question about designation

  1. #1
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    basic question about designation

    I am learning calculus at home during my summer break.In my current lesson, here's the problem:Let 2x squared y + y squared = cos x.Find dy/dx.The solution is dy/dx = -sinx -4xy/2x squared + 2yThe intermediate steps:2x^2 dy/dx + y(4x) + 2y dy/dx = -sin xdy/dx(2x^2 + 2y)= -sin x - 4xyWhy is there dy/dx only for 2x squared and 2y and not -sin x and y(4x)?Any light you could shed on this would be greatly appreciated.Carrie
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  2. #2
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    Hello! Please visit this thread to learn how to write posts using LaTeX: http://www.mathhelpforum.com/math-he...-tutorial.html

    As your post is at the moment, it's very hard to figure out what the question is.
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  3. #3
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    Yes, LaTeX is very fun to use. I think you'll like it.

    The reason we don't get a factor of \frac{dy}{dx} is that the Product Rule states

    \frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x).

    Here, f(x)=2x^2 and g(x)=y. Thus, we have

    \frac{d}{dx}(2x^2\cdot y)=4x\cdot y + 2x^2\cdot\frac{dy}{dx}.
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  4. #4
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    Thank you

    Yes, I see. Thanks for your time, and I"ll check the formatting rules for future use. Please forgive me.
    Carrie
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