# implicit differentiation

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#1
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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4 years ago
#2
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#3
(Original post by High Stakes)
rearrange so y = 6x sub into x + 2y = 0

get 13x = 0?
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4 years ago
#4
(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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#5
(Original post by High Stakes)
Sub y = 6x into your original equation (pre-differential)
Oh ok..
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4 years ago
#6
(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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#7
(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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4 years ago
#8
(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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#9
(Original post by Zacken)
Yes.
Ok thanks, kinda struggle with the applications rather than implicit differentiation itself.
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#10
(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?
0
4 years ago
#11
(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? 0
#12
Thanks.
0
4 years ago
#13
(Original post by SunDun111)
Thanks.
No worries.
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