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Math Help - Implicit Differentiation & Indefinite Integral :(

  1. #1
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    Implicit Differentiation & Indefinite Integral :(

    Help guys - I have loads of these to do and I can't get my head to work. I would really appreciate some detailed help so I can do the rest without exploing

    sin(2x+3y)=3x^3y^2+4
    imp. dif. to get dy/dx=f(x,y)

    and

    find the indefinite integral of g(x)=1 / (x^2 - 25)

    for (-5 < x < 5 )

    Thank you for any help you can give me!
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  2. #2
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    The trick for the first problem is to look at both sides of the equation as functions of x and differentiate them, using the Chain Rule for y.

    This gives us

    \frac{d}{dx}\sin(2x + 3y) = \frac{d}{dx}(3x^3y^2 + 4)
    \cos(2x + 3y)\cdot 3\frac{dy}{dx} = 9x^2y^2 + 3x^3\cdot 2y\frac{dy}{dx}
    \frac{dy}{dx}(\cos(2x + 3y)\cdot3 - 3x^3\cdot2y) = 9x^2y^2
    \frac{dy}{dx} = 3x^2y^2\cdot(\cos(2x + 3y) - 2x^3y)^{-1}.

    For the second problem, it is useful to ask what constants A and B could make the equation

    \frac{1}{(x + 5)(x - 5)} = \frac{A}{x + 5} + \frac{B}{x - 5}

    true. This way, the integrands are simplified. In finding A and B, we use the fact that the equation must hold for all x, not just one value.
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  3. #3
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    Quote Originally Posted by Scott H View Post
    The trick for the first problem is to look at both sides of the equation as functions of x and differentiate them, using the Chain Rule for y.

    This gives us

    \frac{d}{dx}\sin(2x + 3y) = \frac{d}{dx}(3x^3y^2 + 4)
    \cos(2x + 3y)\cdot 3\frac{dy}{dx} = 9x^2y^2 + 3x^3\cdot 2y\frac{dy}{dx}
    \frac{dy}{dx}(\cos(2x + 3y)\cdot3 - 3x^3\cdot2y) = 9x^2y^2
    \frac{dy}{dx} = 3x^2y^2\cdot(\cos(2x + 3y) - 2x^3y)^{-1}.

    For the second problem, it is useful to ask what constants A and B could make the equation

    \frac{1}{(x + 5)(x - 5)} = \frac{A}{x + 5} + \frac{B}{x - 5}

    true. This way, the integrands are simplified. In finding A and B, we use the fact that the equation must hold for all x, not just one value.
    I'm not sure exactly what the question means by getting the answer into dy/dx =f(x,y) Does it just mean that it is a function of x and y in one expression?

    Am I being very dense?!

    Thanks so much for your help; I shall try and go through it at once!
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  4. #4
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    That means it's okay to have y on the other side of the equation.
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  5. #5
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    Quote Originally Posted by Scott H View Post
    That means it's okay to have y on the other side of the equation.
    so your final line of working is in the right format?

    Thanks man

    <tries to apply this to loads more!>
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