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# 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
2. (Original post by High Stakes)
rearrange so y = 6x sub into x + 2y = 0

get 13x = 0?
3. (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)
4. (Original post by High Stakes)
Sub y = 6x into your original equation (pre-differential)
Oh ok..
5. (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
6. (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?
7. (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.
8. (Original post by Zacken)
Yes.
Ok thanks, kinda struggle with the applications rather than implicit differentiation itself.
9. (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?
10. (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?

11. (Original post by Zacken)

Thanks.
12. (Original post by SunDun111)
Thanks.
No worries.

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