Implicit Differentiation

**spoc21** · May 13th 2010, 08:19 AM

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!

**Silver** · May 13th 2010, 08:40 AM

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:

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

**spoc21** · May 13th 2010, 09:16 AM

Originally Posted by Silver

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

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

could you please help me with that? I'm still a little confused.

Thanks!

**Silver** · May 13th 2010, 09:20 AM

Originally Posted by spoc21

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

could you please help me with that? I'm still a little confused.

Thanks!

It shouldn't make a difference

**spoc21** · May 13th 2010, 09:26 AM

Originally Posted by Silver

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

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

is mine incorrect?

**Silver** · May 13th 2010, 09:38 AM

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?

**spoc21** · May 13th 2010, 09:58 AM

yes, makes perfect sense, Thank you!

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