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Math Help - Differentiate using implicit differentiation

  1. #1
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    Differentiate using implicit differentiation

    find dy/dx (3xy)^(1/2)=3x-2y

    Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.



    Answer6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy
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  2. #2
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    Quote Originally Posted by math34 View Post
    find dy/dx (3xy)^(1/2)=3x-2y

    Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.



    Answer6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy
    \sqrt{3xy} = 3x - 2y

    So
    \frac{1}{2} \frac{1}{\sqrt{3xy}} ( 3y + 3xy') = 3 - 2y'

    \frac{3y}{2\sqrt{3xy}} + \frac{3x}{2\sqrt{3xy}}y' = 3 - 2y'

    \frac{3x}{2\sqrt{3xy}}y' + 2y' = 3 - \frac{3y}{2\sqrt{3xy}}

    \left ( \frac{3x}{2\sqrt{3xy}} + 2 \right )y' = 3 -  \frac{3y}{2\sqrt{3xy}}

    y' = \frac{3 -  \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2}

    y' = \frac{3 -  \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2} \cdot \frac{2 \sqrt{3xy}}{2 \sqrt{3xy}}

    y' = \frac{6\sqrt{3xy} - 3y}{3x + 4\sqrt{3xy}} <-- To get your given answer, multiply top and bottom by \sqrt{3xy} again.

    We should rationalize this, but it is easier to use the original equation:
    \sqrt{3xy} = 3x - 2y

    y' = \frac{6(3x - 2y) - 3y}{3x + 4(3x-2y)}

    y' = \frac{18x - 12y - 3y}{3x + 12x - 8y}

    y' = \frac{18x - 15y}{15x - 8y}

    -Dan
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  3. #3
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    Hello, math34!

    Find \frac{dy}{dx}:\;\;(3xy)^{\frac{1}{2}} \:=\:3x-2y

    Would it be easier to first square both sides to get rid of the fractional exponent,
    then differentiate implicitly? . . . absolutely!

    Answer: (6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy . . . this can't be right!

    Square: . 3xy \:=\:9x^2 - 12xy + 4y^2

    We have: . 3xy' + 3y \:=\:18x - 12xy' - 12y + 8yy'

    Then: . 15xy' - 8yy' \:=\:18x - 15y

    Factor: . (15x - 8y)y' \:=\:18x - 15y

    Therefore: . \boxed{y' \:=\:\frac{18x - 15y}{15x - 8y}} . . . which is Dan's answer.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    It can be done head-on . . .

    Differentiate implicitly: . \frac{1}{2}(3xy)^{-\frac{1}{2}}(3y + 3xy') \:=\:3 - 2y'

    We have: . 3y + 3xy' \:=\:2\sqrt{3xy}(3 - 2y')


    We are told that: . \sqrt{3xy} \:=\:3x - 2y

    . . Hence, we have: . 3y + 3xy' \:=\:2(3x - 2y)(3 - 2y')

    Expand: . 3y + 3xy' \:=\:18x - 12xy' - 12y + 8yy'

    Simplify: . 15xy' - 8yy' \:=\:18x - 15y

    Factor: . (15x - 8y)y' \:=\:18x - 15y

    Therefore: . \boxed{y' \:=\:\frac{18x-15y}{15x-8y}}

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  4. #4
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    Thanks

    I got the same answer as you and Dan but, still am clueless about how the given solution was obtained. I think this is a case of the experts making a mistake, while the layman suffers!!!

    Thanks guys
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  5. #5
    Forum Admin topsquark's Avatar
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    I am not happy about the idea of squaring the equation, and here's why:

    Say we have:
    y = x

    Square both sides:
    y^2 = x^2

    Now take the derivative:
    2yy' = 2x

    y' = \frac{x}{y}

    Now, this certainly is a differential equation for the original function, but we also have another solution: y = -x. In this case the differential equation has more solutions than what we started with.

    Just something to watch out for.

    -Dan
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