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Math Help - Implicit differentation help

  1. #1
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    Implicit differentation help

    I have to find the slope of the tangent line to the curve at the point (2, 8):

    √4x+4y + √2xy = 12

    Or also can be written as:

    (4x+4y)^(1/2) + (2xy)^(1/2) = 12

    I tried differentating got got:

    (d/dx)[(4x+4y)^(1/2)] = (1/2)*(4x+4y)^(-1/2)*[4+4(dy/dx)]
    (d/dx)[(2xy)^(1/2)] = (1/2)*(2xy)^(-1/2)*[2y+2x*(dy/dx)]


    then,

    (2+2*(dy/dx))/(4x+4y)^1/2 + (4+x*dy/dx)/(2xy)^(1/2)

    I then plugged in numbers and ended up with (4+4sqrt5)/((-2sqrt5)-4). However that answer is wrong. Can anyone give a clue as to what i did wrong?
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  2. #2
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    Re: Implicit differentation help

    Quote Originally Posted by FrustratedCollegeKid View Post
    I have to find the slope of the tangent line to the curve at the point (2, 8):

    √4x+4y + √2xy = 12

    Or also can be written as:

    (4x+4y)^(1/2) + (2xy)^(1/2) = 12

    I tried differentating got got:

    (d/dx)[(4x+4y)^(1/2)] = (1/2)*(4x+4y)^(-1/2)*[4+4(dy/dx)]
    (d/dx)[(2xy)^(1/2)] = (1/2)*(2xy)^(-1/2)*[2y+2x*(dy/dx)]


    then,

    (2+2*(dy/dx))/(4x+4y)^1/2 + (4+x*dy/dx)/(2xy)^(1/2)

    I then plugged in numbers and ended up with (4+4sqrt5)/((-2sqrt5)-4). However that answer is wrong. Can anyone give a clue as to what i did wrong?
    That last expression should be an equation, plus it has a typo.

    (2+2*(dy/dx))/(4x+4y)^1/2 + (y+x*dy/dx)/(2xy)^(1/2) = 0
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  3. #3
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    Re: Implicit differentation help

    Hello, FrustratedCollegeKid!

    It would help if you told us the (alleged) right answer.

    Also, there must be typo in the problem.
    . . The point (2, 8) is not on the graph.


    Find the slope of the tangent line to the curve at the point (2, 8):

    . . \sqrt{4x+4y} + \sqrt{2xy} \:=\: 12

    We have: . 2(x+y)^{\frac{1}{2}} + \sqrt{2}(xy)^{\frac{1}{2}} \;=\;12

    Then: . 2\cdot\frac{1}{2}(x+y)^{-\frac{1}{2}}\left(1 + \frac{dy}{dx}\right) + \sqrt{2}\cdot\frac{1}{2}(xy)^{-\frac{1}{2}}\left(x\frac{dy}{dx} + 1\cdot y\right) \;\;=\;\;0

    . . \frac{1}{\sqrt{x+y}}\left(1 + \frac{dy}{dx}\right) + \frac{1}{\sqrt{2xy}}\left(x\,\frac{dy}{dx} + y\right) \;\;=\;\;0

    . . \frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{x+y}}\,\frac {dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} + \frac{\sqrt{y}}{\sqrt{2x}} \;\;=\;\;0

    . . \frac{1}{\sqrt{x+y}}\,\frac{dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} \;\;=\;\; -\frac{1}{\sqrt{x+y}} - \frac{\sqrt{y}}{\sqrt{2x}}

    . . \left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{x}}{\sqrt{2y}}\right)\frac{dy}{dx} \;\;=\;\; -\left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{y}}{\sqrt{2x}}\right)

    . . \left(\frac{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}{\sqrt{2y}\sqrt{x+y}}\right) \frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right)

    . . \frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right) \left(\frac{\sqrt{2y}\sqrt{x+y}}{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}\right)

    . . \frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{y}\left(\sqrt{2x} + \sqrt{y}\sqrt{x+y}\right)}{\sqrt{x}\left(\sqrt{2y} + \sqrt{x}\sqrt{x+y}\right)}

    . . \frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{2xy} + y\sqrt{x+y}}{\sqrt{2xy} + x\sqrt{x+y}}

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  4. #4
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    Re: Implicit differentation help

    Quote Originally Posted by Soroban View Post
    Hello, FrustratedCollegeKid!

    It would help if you told us the (alleged) right answer.

    Also, there must be typo in the problem.
    . . The point (2, 8) is not on the graph.


    We have: . 2(x+y)^{\frac{1}{2}} + \sqrt{2}(xy)^{\frac{1}{2}} \;=\;12

    Then: . 2\cdot\frac{1}{2}(x+y)^{-\frac{1}{2}}\left(1 + \frac{dy}{dx}\right) + \sqrt{2}\cdot\frac{1}{2}(xy)^{-\frac{1}{2}}\left(x\frac{dy}{dx} + 1\cdot y\right) \;\;=\;\;0

    . . \frac{1}{\sqrt{x+y}}\left(1 + \frac{dy}{dx}\right) + \frac{1}{\sqrt{2xy}}\left(x\,\frac{dy}{dx} + y\right) \;\;=\;\;0 *

    . . \frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{x+y}}\,\frac {dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} + \frac{\sqrt{y}}{\sqrt{2x}} \;\;=\;\;0

    . . \frac{1}{\sqrt{x+y}}\,\frac{dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} \;\;=\;\; -\frac{1}{\sqrt{x+y}} - \frac{\sqrt{y}}{\sqrt{2x}}

    . . \left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{x}}{\sqrt{2y}}\right)\frac{dy}{dx} \;\;=\;\; -\left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{y}}{\sqrt{2x}}\right)

    . . \left(\frac{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}{\sqrt{2y}\sqrt{x+y}}\right) \frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right)

    . . \frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right) \left(\frac{\sqrt{2y}\sqrt{x+y}}{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}\right)

    . . \frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{y}\left(\sqrt{2x} + \sqrt{y}\sqrt{x+y}\right)}{\sqrt{x}\left(\sqrt{2y} + \sqrt{x}\sqrt{x+y}\right)}

    . . \frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{2xy} + y\sqrt{x+y}}{\sqrt{2xy} + x\sqrt{x+y}}
    As a sidenote, when required to get the value of dy/dx at a given point, I find it much easier NOT to first make dy/dx the subject. Rather, I'd substitute the point into the line marked * and then simplify and make dy/dx the subject. Less algebra, less chance of error, less time taken.
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  5. #5
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    Re: Implicit differentation help

    Hello, mr F!

    Absolutely right . . . excellent advice!
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