# Using chain rule on a square root function

• Dec 9th 2012, 08:18 PM
Chaim
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 :)
• Dec 9th 2012, 08:39 PM
Prove It
Re: Using chain rule on a square root function
The easiest way to remember the Chain Rule is \displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align*}. So when you are trying to differentiate just the part \displaystyle \displaystyle \begin{align*} \sqrt{r^2 - x^2} = \left( r^2 - x^2 \right)^{\frac{1}{2}} \end{align*}, let \displaystyle \displaystyle \begin{align*} u = r^2 - x^2 \end{align*} so that \displaystyle \displaystyle \begin{align*} y = u^{\frac{1}{2}} \end{align*}. Then \displaystyle \displaystyle \begin{align*} \frac{du}{dx} = -2x \end{align*} (which is where the -2x you were wondering about comes from) and \displaystyle \displaystyle \begin{align*} \frac{dy}{du} = \frac{1}{2} \left( r^2 - x^2 \right)^{-\frac{1}{2}} \end{align*}.
• Dec 9th 2012, 09:16 PM
SworD
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.