# Thread: Calculus help for test -- find the detrivative when there is y-prime invovled

1. ## Calculus help for test -- find the detrivative when there is y-prime invovled

Hey guys, its my first post here and I was wondering if you could help.

I need to show the point on the graph of:

x^2-3xy+y^2=1

has a horizontal tangent line.

I am assume that since I know when its horizontal = 0 for slope and y'=0

ps (when I am writing y' it = yprime)

So step one is find the derivative.

The teacher told us the derivative is 2x-3y-3xy'-2yy'=0

is this right ? can anyone explain it?

2. Originally Posted by richtree
Hey guys, its my first post here and I was wondering if you could help.

I need to show the point on the graph of:

x^2-3xy+y^2=1

has a horizontal tangent line.

I am assume that since I know when its horizontal = 0 for slope and y'=0

ps (when I am writing y' it = yprime)

So step one is find the derivative.

The teacher told us the derivative is 2x-3y-yxy'-2yy'=0

is this right ? can anyone explain it?
So 2x- 3y- (yx+ 2y)y'= 0. Solve for y'.

3. hallsof ivy, thanks for responding,

but my teacher is telling me the derivative is :

2x-3y-3xy'-2yy'=0

I don't know how this came from the original equation ?

4. x^2-3xy+y^2=1
you are finding an implicit derivative w/r to x ... treat y as a function of x

$\displaystyle \frac{d}{dx} (x^2 - 3xy + y^2 = 1)$

don't forget the product rule for the middle term ...

$\displaystyle 2x -3x \frac{dy}{dx} - 3y + 2y\frac{dy}{dx} = 0$

isolate $\displaystyle \frac{dy}{dx}$ ...

$\displaystyle -3x \frac{dy}{dx} + 2y\frac{dy}{dx} = 3y - 2x$

$\displaystyle \frac{dy}{dx}(2y - 3x) = 3y - 2x$

$\displaystyle \frac{dy}{dx} = \frac{3y-2x}{2y-3x}$

5. thanks to both of you, I still just don't understand it, the teacher really confused me by not showing the rational explanation that the book uses.

If anyone could do a step by step solution with text I would appreciate it.

I think if I could see the whole problem start to finish maybe I could grasp it better

6. ok, skeeter, I'm with so far far, can you continue?

thx guys

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ok, so y = 2/3x

now I take 2/3x and plug it into the original equation.

So,

x^2-3xy+y^2=1

becomes

x^2-3x(2/3x) + (2/3x)2 = 1

right??????? I think I am close . help me finish !!!!

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so next is

x^2 - 2x^2 +4/9x^2 = 1

then what ???????????

I'm offically stuck !!!!!

7. Originally Posted by richtree
so next is

x^2 - 2x^2 +4/9x^2 = 1

then what ???????????

I'm offically stuck !!!!!
$\displaystyle x^2 - 3x\left(\frac{2x}{3}\right) + \left(\frac{2x}{3}\right)^2 = 1$

$\displaystyle x^2 - 2x^2 + \frac{4x^2}{9} = 1$

combine like terms ...

$\displaystyle -\frac{5x^2}{9} = 1$

$\displaystyle x^2 = -\frac{9}{5}$

this equation has no real solution ... so, what does that tell you?

8. yeah, so x= pos.neg.sqareroot of -9/5

so there is no point right ?

9. Originally Posted by richtree
yeah, so x= pos.neg.sqareroot of -9/5

so there is no point right ?
correct ... no horizontal tangent.