Implicit Differentiation

1. Feb 24, 2009

Mentallic

I have been able to follow how to take the derivative of implicit functions, such as:

$$x^2+y^2-1=0$$

Differentiating with respect to x

$$2x+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=\frac{-x}{y}$$

Sure it's simple to follow, but I don't understand why the $$\frac{dy}{dx}$$ is tacked onto the end of the differentiated variable y.

An explanation or article on the subject would be appreciated. Thanks.

2. Feb 24, 2009

cristo

Staff Emeritus
You're differentiating wrt x, so using the chain rule:

$$\frac{d}{dx}(y^2)=\frac{d}{dy}(y^2)\frac{dy}{dx}=2y\frac{dy}{dx}$$

3. Feb 24, 2009

Mentallic

Aha, so it's done using the chain rule. Thankyou

4. Feb 24, 2009

HallsofIvy

Staff Emeritus
It's not "tacked on", it's nailed firmly!:tongue2:

5. Feb 25, 2009

Mentallic

haha :rofl:
I always think 2 moves ahead, taking into consideration that separating to isolate will be necessary. Nail vs tack, I think we know the winner

Last edited: Feb 25, 2009