How to take the derivative of implicit functions

[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.

 

[cristo]

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

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

 

[Mentallic]

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

 

[HallsofIvy]

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

 

[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