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elleblythe

  • one year ago

Find the dy/dx of: 6x^(4/3) + 2y^5 = x^3y^2 + cube root of pi^5

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  1. Loser66
    • one year ago
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    where are you stuck?

  2. elleblythe
    • one year ago
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    @Loser66 is the correct answer (not simplified): -cube root of 8 / (-10y^4 + 2x^3y + 3x^2y) ?

  3. Loser66
    • one year ago
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    You guessed? or just pick one of the options and check it from me?

  4. Loser66
    • one year ago
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    Show me your work, please. Just take derivative both sides and isolate y'. Done.

  5. elleblythe
    • one year ago
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    @Loser66 I solved for it. \[8x ^{1/3}+10y ^{4}(dy/dx)=x^32y(dy/dx)+3x^2y^2(dy/dx)+5/3\pi^^{2/3}\]

  6. Loser66
    • one year ago
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    the left hand side is ok, but the right one!! first term of the right one is \(x^3y^2\), hence its derivative is \(3x^2y^2+2x^3y (dy/dx)\) the last term is a constant, hence its derivative =0 , ignore it

  7. Loser66
    • one year ago
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    now, combine and isolate dy/dx, please

  8. elleblythe
    • one year ago
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    @Loser66 -cube root of 8 + 3x^2y^2 / (-10y^4 + 2x^3y)

  9. Loser66
    • one year ago
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    the sign of denominator is not correct, it should be \(\dfrac{dy}{dx}=\dfrac{3x^2y^2-\sqrt[3]{x}}{10y^4-2x^3y}\)

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