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Math Help - Differentiating Implicit Equations

  1. #1
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    Angry Differentiating Implicit Equations

    If
    Code:
    sin^4(\[Alpha])/x+cos^4(\[Alpha])/y-1/(x+y)==0
    ^^^^(it looked fine in mathematica...can someone tell me how to write code in this forum?)

    then what is
    Code:
    dy/dy??
    *here [alpha] is a constant*


    well i figured out dy/dx and it came out to be
    Code:
    y/x
    but the book says
    Code:
    dy/dx=tan^2[Alpha]
    ,which is a constant. Does this mean the above line represents a straight line??Mathematica plotted a pair of lines(i guess) and i cannot ask mathematica to find
    Code:
     dy/dx
    (if its possible to find (dy/dx) from implicit equations in mathematica then tell me!)
    Thanks!
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  2. #2
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    Re: Differentiating Implicit Equations

    Hello, smatik!

    I don't understand your equation.
    A few SPACES would be welcome.



    sin^4(\[Alpha])/x+cos^4(\[Alpha])/y-1/(x+y)==0

    You seem to have: . \sin^4\alpha \div x + \cos^4\alpha \div y - 1 \div(x+y)\:==\: 0


    I see several interpretations for this . . .

    . . \frac{\sin^4\alpha}{x} + \frac{\cos^4\alpha}{y} - \frac{1}{x+y} \:=\:0

    . . \sin^4\left(\frac{\alpha}{x}\right) + \cos^4\left(\frac{\alpha}{y}\right) - \frac{1}{x+y} \:=\:0

    . . and so on.
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  3. #3
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    Re: Differentiating Implicit Equations

    The first one is it!!!! How to write like you did in this forum??
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  4. #4
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    Re: Differentiating Implicit Equations

    If you hover your mouse over a math expression rendered in latex on this forum you can see the latex code for that expression.

    To render the code, surround your code in [tex][/tex] tags, for example:

    [tex]\frac{2}{3\cos\alpha}[/tex] gives:

    \frac{2}{3\cos\alpha}
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  5. #5
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    Re: Differentiating Implicit Equations

    Thanks.I meant this this.....
    \frac{sin^4(\alpha)}{x}+\frac{cos^4(\alpha)}{x}-\frac{1}{x+y} = 0
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  6. #6
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    Re: Differentiating Implicit Equations

    About the dy/dx thing. I haven't actually checked, but I believe that the Mathematica uses the form "d/dx(...)" for derivatives. Technically speaking this is probably a partial derivative.

    -Dan
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  7. #7
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    Re: Differentiating Implicit Equations

    Hello, smatik!

    \text{Differentiate: }\:\frac{\sin^4\!\alpha}{x}+\frac{\cos^4\!\alpha}{  y}-\frac{1}{x+y} \:=\: 0

    Since \alpha is a constant,
    . . let: a \,=\,\sin^4\!\alpha,\;b \,=\,\cos^4\!\alpha

    We have: . ax^{\text{-}1} + by^{\text{-}1} - (x+y)^{\text{-}1} \:=\:0

    Then: . -ax^{\text{-}2} - by^{\text{-}2}y' - (x+y)^{\text{-}2}(1+y') \:=\:0

    . . . . . -\frac{a}{x^2} - \frac{b}{y^2}y' - \frac{1}{(x+y)^2} - \frac{1}{(x+y)^2}y' \:=\:0


    Multiply by x^2y^2(x+y)^2\!:

    . . -ay^2(x+y)^2 - bx^2(x+y)^2y' - x^2y^2 - x^2y^2y' \:=\:0

    . . . . -bx^2(x+y)^2y' - x^2y^2y' \:=\:ay^2(x+y)^2 + x^2y^2

    . . . . -x^2\left[b(x+y)^2 + y^2\right]y' \:=\:y^2\left[a(x+y)^2 + x^2\right]

    . . . . . . . . . . . y' \;=\;-\frac{y^2\left[a(x+y)^2 + x^2\right]}{x^2\left[b(x+y)^2 + y^2\right]}

    . . . . . . . . . . . y' \;=\;-\frac{y^2\left[(x+y)^2\sin^4\!\alpha + x^2\right]}{x^2\left[(x+y)^2\cos^4\!\alpha + y^2\right]}
    Thanks from smatik
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