# Thread: dy dx of the curve

1. ## dy dx of the curve

find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Cheers

2. Originally Posted by Joel
find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Cheers
first of all, it's dy/dx , not dydx.

second, you're not even close in your "method".

research finding implicit derivatives in your text or on the web.

3. Just to clarify a bit on what skeeter is saying:

To use the more "traditional" method of solving for a derivative, you would need to solve for y explicitly. What that means is that you can get the equation into the form $y=\cdots$ where there are no $y$ terms on the right hand side at all.

Although your work to solve for $y$ seemed like it was on the right track, the problem is you are stuck with a $y$ term in the right hand side.

The correct way to handle this problem is to use implicit differentiation. This method is used without first solving for $y$.

Just to give you a small hint -- You do not need to simply or modify the equation at all before taking the derivative! Have you learned implicit differentiation techniques?

4. Originally Posted by Joel
find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Cheers
When you have $f(y)$, it is the same as evaluating $f(x)$ except that there is a $\frac{dy}{dx}$ attached (multiplied).

So deriving $y^2 = x$ gives

$2y\frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{2y}$

This is simply a special case of the chain rule: notice that when evaluating x's this rule is hidden by the fact $\frac{dx}{dx} = 1$.

Product rules with $f(y)g(x)$ work as you would expect. Can you use this technique to answer your question?