# dy dx of the curve

#### 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

#### skeeter

MHF Helper
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.

#### drumist

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 $$\displaystyle y=\cdots$$ where there are no $$\displaystyle y$$ terms on the right hand side at all.

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

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?

#### Turiski

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

So deriving $$\displaystyle y^2 = x$$ gives

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

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

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

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