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implicit differentiation

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
Reply 2
Original post by High Stakes

rearrange so y = 6x sub into x + 2y = 0

get 13x = 0?
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)
Reply 4
Original post by High Stakes
Sub y = 6x into your original equation (pre-differential)


Oh ok..
Reply 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
Reply 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?
Reply 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.
Reply 8
Original post by Zacken
Yes.


Ok thanks, kinda struggle with the applications rather than implicit differentiation itself.
Reply 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?
Reply 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?


It's the same answer:

3x2siny+y2×y2y2=3x2y2y2siny+1\displaystyle \frac{3x^2}{\sin y + y^{-2}} \times \frac{y^2}{y^2} = \frac{3x^2 y^2}{y^2 \sin y + 1}
Reply 11
Original post by Zacken
It's the same answer:

3x2siny+y2×y2y2=3x2y2y2siny+1\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.
Reply 12
Original post by SunDun111
Thanks.


No worries.

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