# Quick derivative question

1. Sep 16, 2007

### fitz_calc

Where does this 2x (highlighted) come from? I thought this step was just the 'product rule', but it looks like the 'chain rule' was applied as (x^2 + 2)`. Were these two rules used simultaneously?

2. Sep 16, 2007

### Dick

First they used the product rule, and then they used the chain rule to find the derivative of sqrt(x^2+2). The 2x comes from the derivative of x^2+2.

3. Sep 17, 2007

### HallsofIvy

Staff Emeritus
That is actually a double application of the chain rule. To differentiate $\sqrt{x^2+1}$, let u= x2 and let v= u+ 1. $\sqrt{x^2+ 1}$ becomes $\sqrt{v}= v^{1/2}$. It's derivative is (1/2)v-1/2v'. Of course, v'= u'= 2x.

(I am ignoring the x multipying the square root since you only asked about the chain rule.)