Using chain rule on a square root function

An example that my teacher did was:

y=2x(√(r^{2}-x^{2})

Then the derivative would be:

y'=2x((1/2)(r^{2}-x^{2})^{(-1/2)}(-2x))+(r^{2}-x^{2})^{(1/2)}(2)

So what I understand is that he took the derivative inside, then added the derivative of the outside times inside

But I'm confused where he got the -2x from, inside the parenthesis,** can someone explain?**

Chain rule and product rule is kind of my worst, __does anyone know a good and simple rule to follow for chain rule?__

or someway to kind of explain the chain rule a bit, because it's kind of confusing, especially with a fraction exponent.

Like for example, what I use for the quotient rule is (low d high) - (high d low) over the denominator square it goes.

So it's denominator * derivative of numerator - numerator * derivative of denominator, divided by denominator squared.

So if anyone know a good rhyme or a good simple way to remember the chain rule, and product rule as well, think you can share it to me as well? Thanks :)

Re: Using chain rule on a square root function

The easiest way to remember the Chain Rule is . So when you are trying to differentiate just the part , let so that . Then (which is where the -2x you were wondering about comes from) and .

Re: Using chain rule on a square root function

For the product rule, just remember that each of the two parts "takes a turn" being differentiated while the other is left alone. Furthermore its just like the quotient rule except the negative sign is a positive sign and there is no division by g^2.