# How to take the derivative of implicit functions

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

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