Understanding the Derivative: Product Rule vs. Chain Rule

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fitz_calc
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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?

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That is actually a double application of the chain rule. To differentiate [itex]\sqrt{x^2+1}[/itex], let u= x2 and let v= u+ 1. [itex]\sqrt{x^2+ 1}[/itex] becomes [itex]\sqrt{v}= v^{1/2}[/itex]. 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.)