Homework Help: Differentiating x² + 2xy - 3y² + 16 = 0

Differentiate Implicitly.

You should get [itex]\mathrm{d}y/ \mathrm{d}x[/itex] within the differentiated equation twice. Factorise [itex]\mathrm{d}y/ \mathrm{d}x[/itex] (by taking it to one side) and rearrange it so that it is only on one side. Then solve for co-ordinates.

 

We should write that first line as

d/dx (x² + 2xy - 3y² + 16) = d/dx (0)

i.e., it's "d/dx" and not "dy/dx" in front, since we want to take the derivative with respect to x.

Then use the chain rule, and when needed the product rule. For example:

[tex]
\frac{d}{dx}(x^2 y^4)
= 2x \cdot y^4 + x^2 \cdot 4 y^3 \frac{dy}{dx}
[/tex]

 

Let's suppose u = y^4, and we want to differentiate that with respect to x.

d/dx (y^4)
= d/dx (u)
= du/dx
and by the chain rule
= (du/dy) * (dy/dx)
= 4 y^3 * (dy/dx)

Moreover:

d/dx (2xy) is not 2y. You need the product rule here, applied to the functions "x" and "y".

 

Hi thomas49th!

Simple question:

Simple answer:

y is 3y², and u is y.

oh yes they are …

d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx.

 

Hi thomas49th!

Your original problem was implicit differentiation of x² + 2xy - 3y² + 16:

that is, d/dx of x² + 2xy - 3y² + 16.

Obviously , d/dx of x² + 16 is really easy, and d/dx of 2xy is fairly easy … I assume you can do that …

So that leaves d/dx of 3y².

You know how to do d/dy of 3y².

So you use the chain rule: d(3y²)/dx = d(3y²)/dy dy/dx = 6y dy/dx.

Does that make sense?

 

thomas the idea is really simple.

if you have y^2 then different as if it was x , you should have 2y. simple right?

Now all you have to do is multiply the 2y by dy/dx. Just write 2y . dy/dx and that's the answer.

So whenever you different with the letter y, do it as if it was x but just multiply it by dy/dx. This is because of the chain rule, since y is a function of x.

 

d/dx of 2xy

= d(2xy)/dx = yd(2x)/dx + 2x dy/dx = 2y + 2x dy/dx.

 

You're right … it's definitely the first one.

(2y . (dy/dx) would be the d/dx of y²)

Then, as you said in your first post:

So just put dy/dx = 0 in 2x + 2y + 2x . (dy/dx) - 6y.dy/dx

that gives you … ?

(going to bed now … :zzz:)

 

Not quite but you are really close.

2x + 2y = 0
Divide both sides of equation by 2:
x + y = 0
And then
x = ???

 

No. It is NOT right.

2x+ 2y= 0 is right. But you solved that equation wrong!

 

… yawn … stretch … rubs eyes …

It's right!!

well … apart from the minus sign!!

Now put y = -x into the original curve equation x² + 2xy - 3y² + 16 = 0, to find the points where dy/dx = 0, as required.

 

Hi thomas49th!

erm … you're missing the obvious … you have x = ±2, and x = -y.

So you don't need to put y back into the equation, do you?