Using chain rule on a square root function
An example that my teacher did was:
y=2x(√(r2-x2)
Then the derivative would be:
y'=2x((1/2)(r2-x2)(-1/2)(-2x))+(r2-x2)(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.