SA3Why does differentiating y^2 give 2yy' when using implicit differentiation?

[vanmaiden]

differentiating for "y"

Homework Statement

I don't quite understand the reason why one can \frac{d}{dx} y² and get 2y \frac{dy}{dx} when using implicit differentiation. Why can one even go as far as to change y² to 2y?

Homework Equations

The Attempt at a Solution

I see the chain rule is applied, I just would like to have a good understanding as to why differentiating y² gives 2y \frac{dy}{dx} when using implicit differentiation

 

[wisvuze]

you should think of it as y = f ( x ), as y is a function of x.
Then, the use of the chain-rule is obvious. Suppose you have the function T ( x ) = 2 + [ f( x ) ]^2 = 2 + y^2 . Then if you want to differentiate this with respect to x, you have to differentiate T ' ( x ) = 2f(x )*f ' ( x )

 

[Ray Vickson]

Homework Statement

I don't quite understand the reason why one can \frac{d}{dx} y² and get 2y \frac{dy}{dx} when using implicit differentiation. Why can one even go as far as to change y² to 2y?

Homework Equations

The Attempt at a Solution

I see the chain rule is applied, I just would like to have a good understanding as to why differentiating y² gives 2y \frac{dy}{dx} when using implicit differentiation

For f(x) = [y(x)]^2 and small h > 0 we have f(x+h) - f(x) = [y(x+h)]^2 - [y(x)]^2. We also have [y(x+h) -y(x)]/h ~ y'(x) for small h > 0 (becoming exact in the limit h --> 0), so f(x+h) - f(X) is almost equal to [y(x) + h*y'(x)]^2 - [y(x)]^2 = 2*y(x)*y'(x)*h + h^2*y'(x)^2, with equality becoming exact in the limit. So, what is limit_{h->0} [f(x+h) - f(x)]/h?

RGV