SunDun111
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The question is the curve has equation xy = 3x^2 - y^2 + 3

Find in surd form the x coordinates of the point P and Q on the cruve at which Dy/Dx = 0

I used implicit differentation and got it to DY/DX = 6x - y/ x + 2y

Yet when i make it equal to 0 i get x and y as 0 which is wrong? any ideas
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High Stakes
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SunDun111
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(Original post by High Stakes)
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rearrange so y = 6x sub into x + 2y = 0

get 13x = 0?
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High Stakes
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(Original post by SunDun111)
rearrange so y = 6x sub into x + 2y = 0

get 13x = 0?
Sub y = 6x into your original equation (pre-differential)
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SunDun111
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(Original post by High Stakes)
Sub y = 6x into your original equation (pre-differential)
Oh ok..
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Qcomber
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(Original post by SunDun111)
The question is the curve has equation xy = 3x^2 - y^2 + 3

Find in surd form the x coordinates of the point P and Q on the cruve at which Dy/Dx = 0

I used implicit differentation and got it to DY/DX = 6x - y/ x + 2y

Yet when i make it equal to 0 i get x and y as 0 which is wrong? any ideas
Once you get dy/dx and set it equal to zero, you get
6x-y/x+2y = 0
so 6x-y =0 (multiply both sides by denominator)
so y = 6x
sub this into the original equation
x(6x) = 3x^2 -(6x)^2 + 3
solve you get to 39x^2 = 3
so x = +- root13/13
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SunDun111
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(Original post by Qcomber)
Once you get dy/dx and set it equal to zero, you get
6x-y/x+2y = 0
so 6x-y =0 (multiply both sides by denominator)
so y = 6x
sub this into the original equation
x(6x) = 3x^2 -(6x)^2 + 3
solve you get to 39x^2 = 3
so x = +- root13/13
thanks i did it, when it comes to stationary points, I have the equation DY/DX = 2(x+y-1)/1-2(x+y), to find the stionary pooint do i set it equal to = 0 then sub it back into the original with a simultaneus equation?
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Zacken
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(Original post by SunDun111)
thanks i did it, when it comes to stationary points, I have the equation DY/DX = 2(x+y-1)/1-2(x+y), to find the stionary pooint do i set it equal to = 0 then sub it back into the original with a simultaneus equation?
Yes.
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SunDun111
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(Original post by Zacken)
Yes.
Ok thanks, kinda struggle with the applications rather than implicit differentiation itself.
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SunDun111
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(Original post by Zacken)
Yes.
Have another question, the curve has equation 1/y + cos y = 5 - x^3, if i implicitly differentiate this i would get dy/dx = 3x^2 / siny + y^-2,

The answer is DY/DX = 3x^2 multiplied by y^2 / 1 + y^2 sin y where have i gone wrong?
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Zacken
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(Original post by SunDun111)
Have another question, the curve has equation 1/y + cos y = 5 - x^3, if i implicitly differentiate this i would get dy/dx = 3x^2 / siny + y^-2,

The answer is DY/DX = 3x^2 multiplied by y^2 / 1 + y^2 sin y where have i gone wrong?
It's the same answer:

\displaystyle \frac{3x^2}{\sin y + y^{-2}} \times \frac{y^2}{y^2} = \frac{3x^2 y^2}{y^2 \sin y + 1}
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SunDun111
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(Original post by Zacken)
It's the same answer:

\displaystyle \frac{3x^2}{\sin y + y^{-2}} \times \frac{y^2}{y^2} = \frac{3x^2 y^2}{y^2 \sin y + 1}
Thanks.
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Zacken
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(Original post by SunDun111)
Thanks.
No worries.
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