1. ## Implicit Differentiation

Use implicit differentiation to find dy/dx for xy2 - yx2 = 3xy.

My working:

$x.(2y.dy/dx)+y^2(1)-y(2x)+dy/dx(x^2)=3x dy/dx + 3y$

$==> 2xy dy/dx + y^2 - 2xy + dy/dx (x^2) = 3x dy/dx + 3y$

Now Im very confused. Do we need to divide both sides, in order to separate dy/dx? Any helpful tips/suggestions would be appreciated

Thanks!

2. So $xy^2 - x^2y = 3xy$

Your working looks right although it's a little messy and I think youre dropping a sign somewhere so allow me:

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

If we throw $\frac{dy}{dx}$'s to one side and the rest to the other side:

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

Take out $\frac{dy}{dx}$ as factor on RHS

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

Divide both sides by $(3x - 2xy + x^2)$ and you end up with:

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

3. Originally Posted by Silver
So $xy^2 - x^2y = 3xy$
Im sorry, I didn't type the question in proper notation. it's actually:
$xy^2-yx^2=3xy$

Thanks!

4. Originally Posted by spoc21
Im sorry, I didn't type the question in proper notation. it's actually:
$xy^2-yx^2=3xy$

Thanks!
It shouldn't make a difference

5. Originally Posted by Silver

$y^2 + 2xy\frac{dy}{dx} - x^2\frac{dy}{dx} - 2xy = 3x\frac{dy}{dx} + 3y$
I actually got $+ x^2\frac{dy}{dx}$, using the product rule/chain rule of derivative.

is mine incorrect?

6. That was the sign I mentioned you dropped earlier.

Look at it this way:

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

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

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

Does that make sense?

7. yes, makes perfect sense, Thank you!