# Implicit Differentiation

#### spoc21

[FONT=&quot]Use implicit differentiation to find [/FONT][FONT=&quot]dy/dx for xy2 - yx2 = 3xy.

My working:[/FONT]
$$\displaystyle x.(2y.dy/dx)+y^2(1)-y(2x)+dy/dx(x^2)=3x dy/dx + 3y$$

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

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! [FONT=&quot]
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#### Silver

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

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

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

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

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

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

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

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

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

#### spoc21

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

Thanks! #### Silver

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

Thanks! It shouldn't make a difference #### spoc21

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

is mine incorrect?

#### Silver

That was the sign I mentioned you dropped earlier.

Look at it this way:

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

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

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

Does that make sense?

• spoc21

#### spoc21

yes, makes perfect sense, Thank you! 