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Math Help - Find the derivative y′(x) implicitly

  1. #1
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    Find the derivative y′(x) implicitly

    Hi!

    \text{A.}\,3xy^3-4x = 10y^2

    A little help with solving this please. I'll happily accept tips on solving implicit problems, considering I am weak at them.
    Thank you!

    3y^3 + x\cdot\frac{d}{dx}\left(y^3\right) + y^3\frac{d}{dx}(x) -\frac{d}{dx}(4x) = 20yy\prime

    3y^3+3xy^2y\prime + y^3 -4 = 20yy\prime

    4y^3+3xy^2y\prime - 4 = 20yy\prime

    y\prime(3xy^2-20y)= 4-4y^3

    y\prime = \frac{4-4y^3}{3xy^2-20y}
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  2. #2
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    Re: Find the derivative y′(x) implicitly

    \left[3x\cdot3y^2y\prime+y^3\cdot3\right]-4 = 20yy\prime
    9xy^2y\prime +3y^3 - 4 = 20yy\prime
    y\prime= \frac{4-3y^3}{9xy^2-20y}
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  3. #3
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    Re: Find the derivative y′(x) implicitly

    Quote Originally Posted by Unreal View Post
    Hi!

    \text{A.}\,3xy^3-4x = 10y^2

    A little help with solving this please. I'll happily accept tips on solving implicit problems, considering I am weak at them.
    Thank you!

    3y^3 + x\cdot\frac{d}{dx}\left(y^3\right) + y^3\frac{d}{dx}(x) -\frac{d}{dx}(4x) = 20yy\prime
    the derivative, with respect to x, of 3xy^3 is 3y^3+ 3x(y^3)'= 3y^3+ 9xy^2y'. You do not have the "3" that was in the original term in the second part.
    I don't know where that " y^3\frac{d}{dx}(x) comes from. You had already done that in first "3y^3[/tex]". The derivative, with respect to x is, in fact -4 so the left side is
    3y^3+ 9xy^2y'- 4. The right is 20yy' as you say so you have 3y^3+ 9xy^2y'- 4= 20yy'

    Subtract 10yy' and 3y^3- 4 from both sides to get 9xy^2y'- 20yy'= (9xy^2- 20y)y'= 4- 3y^3
    Your only error appears to be that missing "3".

    3y^3+3xy^2y\prime + y^3 -4 = 20yy\prime

    4y^3+3xy^2y\prime - 4 = 20yy\prime

    y\prime(3xy^2-20y)= 4-4y^3

    y\prime = \frac{4-4y^3}{3xy^2-20y}
    Thanks from Unreal
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