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Thread: 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 $\displaystyle x$ and differentiate them, using the Chain Rule for $\displaystyle y$.

    This gives us

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

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

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

    true. This way, the integrands are simplified. In finding $\displaystyle A$ and $\displaystyle B$, we use the fact that the equation must hold for all $\displaystyle x$, 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 $\displaystyle x$ and differentiate them, using the Chain Rule for $\displaystyle y$.

    This gives us

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

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

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

    true. This way, the integrands are simplified. In finding $\displaystyle A$ and $\displaystyle B$, we use the fact that the equation must hold for all $\displaystyle x$, 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 $\displaystyle y$ 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 $\displaystyle y$ 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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