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Math Help - 2 simple problems concerning implicit differentiation

  1. #1
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    2 simple problems concerning implicit differentiation

    1) Find dy/dx for 4y^2 - xy = 2

    The given answer is; y / (8y - x) but I don't understand how to get that. My procedure was;

    4y^2 - xy = 2
    8y - (x + y) = 0
    8y - x - y = 0
    And according to how the answer was gotten, that means I should move over the -y to the other side, followed by the 8y - x together. What I don't understand is why you can't subtract the 8y and -y first to make -7y and then move it over? And why is the 8y - x supposed to be moved together to the other side instead of separately?

    2) Find dy/dx for (x + y) / (2x - y) = 1

    I'm assuming you have to apply the quotient rule first, but can anyone tell me how it's supposed to look after the rule's been applied? I'm doing something wrong in my calculations.
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    Quote Originally Posted by Archduke01 View Post
    1) Find dy/dx for 4y^2 - xy = 2

    The given answer is; y / (8y - x) but I don't understand how to get that. My procedure was;

    4y^2 - xy = 2
    8y - (x + y) = 0
    8y - x - y = 0
    And according to how the answer was gotten, that means I should move over the -y to the other side, followed by the 8y - x together. What I don't understand is why you can't subtract the 8y and -y first to make -7y and then move it over? And why is the 8y - x supposed to be moved together to the other side instead of separately?

    2) Find dy/dx for (x + y) / (2x - y) = 1

    I'm assuming you have to apply the quotient rule first, but can anyone tell me how it's supposed to look after the rule's been applied? I'm doing something wrong in my calculations.
    You have to use the chain rule to differentiate. The derivative of 4y^2 is 8yy' because you're differentiating with respect to x. You wrote 8y, but this is just the derivative of 4y^2 with respect to y. So your equation should have been:

    8yy'-(y+xy')=0

    8yy'-y-xy'=0

    y'(8y-x)=y

    y'=\frac{y}{8y-x}
    Last edited by adkinsjr; November 14th 2009 at 02:13 PM.
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    Quote Originally Posted by Archduke01 View Post

    2) Find dy/dx for (x + y) / (2x - y) = 1

    I'm assuming you have to apply the quotient rule first, but can anyone tell me how it's supposed to look after the rule's been applied? I'm doing something wrong in my calculations.
    Yes, you just have to apply the quotient rule. This is how the rule is applied:

    \frac{(2x-y)\frac{d}{dx}(x+y)-(x+y)\frac{d}{dx}(2x-y)}{(2x-y)^2}=0

    Now you just find the two derivatives in the numerator, and substitute them.
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    Quote Originally Posted by adkinsjr View Post
    Yes, you just have to apply the quotient rule. This is how the rule is applied:

    \frac{(2x-y)\frac{d}{dx}(x+y)-(x+y)\frac{d}{dx}(2x-y)}{(2x-y)^2}=0

    Now you just find the two derivatives in the numerator, and substitute them.
    Am I on the right path if I get (2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0 ?
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    Quote Originally Posted by Archduke01 View Post
    Am I on the right path if I get (2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0 ?
    Yup
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    (2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0

    I don't know how to proceed from there. I expanded the terms and ended up with ...

    2x + 2xy' - y - yy' - (2x - xy' + 2y - yy') / (2x - y)^2 = 0
    2xy' + 3xy' - 3y / (2x-y)^2 = 0

    Seems like I messed up somewhere along the line. Could someone please point it out?
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  7. #7
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    Quote Originally Posted by Archduke01 View Post
    (2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0

    I don't know how to proceed from there. I expanded the terms and ended up with ...

    2x + 2xy' - y - yy' - (2x - xy' + 2y - yy') / (2x - y)^2 = 0
    2xy' + 3xy' - 3y / (2x-y)^2 = 0

    Seems like I messed up somewhere along the line. Could someone please point it out?
    Just set the numerator equal to zero, and then expand. You don't need to worry about the denominator. Remember that the denominator can't be
    zero. Just solve this for y':

    (2x - y)(1 + y') - (x+y)(2-y')=0
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    Quote Originally Posted by adkinsjr View Post
    Just set the numerator equal to zero, and then expand. You don't need to worry about the denominator. Remember that the denominator can't be
    zero. Just solve this for y':

    (2x - y)(1 + y') - (x+y)(2-y')=0
    Thanks but I know, that's what I did. I ended up with 2xy' + 3xy' - 3y = 0
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    Quote Originally Posted by Archduke01 View Post
    Thanks but I know, that's what I did. I ended up with 2xy' + 3xy' - 3y = 0

    (2x - y)(1 + y') - (x+y)(2-y')=0

    2x-y+2xy'-yy'-[2x-xy'+2y-yy']=0

    -y+2xy'-2y+xy'=0

    3xy'-3y=0

    y'=\frac{y}{x}

    You made a simple algebraic error, but you were close.
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    Quote Originally Posted by adkinsjr View Post
    (2x - y)(1 + y') - (x+y)(2-y')=0

    2x-y+2xy'-yy'-[2x-xy'+2y-yy']=0

    -y+2xy'-2y+xy'=0

    3xy'-3y=0

    y'=\frac{y}{x}

    You made a simple algebraic error, but you were close.
    Thanks, I corrected it and also got
    3xy'-3y=0 but the thing is, the given answer is 1/2.
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    Quote Originally Posted by Archduke01 View Post
    Thanks, I corrected it and also got
    3xy'-3y=0 but the thing is, the given answer is 1/2.
    I reviewed the problem, I can't see anything wrong with it. Did you state the entire problem in your original post? Perhaps you were asked to find \frac{dy}{dx} at the point (2,1)
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    Quote Originally Posted by adkinsjr View Post
    I reviewed the problem, I can't see anything wrong with it. Did you state the entire problem in your original post? Perhaps you were asked to find \frac{dy}{dx} at the point (2,1)
    Nope. It only asked to find dy/dx. Maybe we were supposed to do something with the denominator?
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  13. #13
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    Wait a minute, lol, hold on!

    \frac{x+y}{2x-y}=1

    x+y=2x-y

    Forget the quotient rule!!!

    1+y'=2-y'

    y'=\frac{1}{2}

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  14. #14
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    Quote Originally Posted by adkinsjr View Post
    Wait a minute, lol, hold on!

    \frac{x+y}{2x-y}=1

    x+y=2x-y

    Forget the quotient rule!!!

    1+y'=2-y'

    y'=\frac{1}{2}

    Oh damn LOL it was as simple as that!

    Anyway, thank you so much for your time and sorry for all the questions.
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