While working on the problem $g(x)= \sqrt{x^{2}-4x+4}$, I got stuck and plugged the formula into Wolfram Alpha. It simplified alot of things for me, but there is one section I'm having trouble with, since one of my weaknesses is radicals.

According to the Chain Rule, $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$, and therefore $u= x^{2}-4x+4$ and $y= \sqrt{u}$, resulting in $\frac{d\sqrt{u}}{du}$. That then transforms to $\frac{1}{2\sqrt{u}}$.

How does that work?

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    $\begingroup$ $\sqrt{f(x)}=[f(x)]^\frac{1}{2}$; so to differentiate this, just apply the rules of differentiation, namely the power rule and the chain rule. $\endgroup$ – Matt Calhoun Feb 23 '11 at 20:20

I sometimes find that working with abstract functions is easier than specific ones, especially when powers and radicals are involved as they are very confusing for me.

The chain rule, as you quote it, is correct. Your implementation, however, is not.

Define $f(u) = \sqrt u$ and $g(x) = x^2 -4x +4$, then you wish to derive $f(g(x))$, which should give $$(f(g(x))' = f'(g(x))\cdot g'(x)=\frac{1}{2\sqrt{g(x)}}\cdot g'(x) = \frac{1}{2\sqrt{x^2-4x+4}}\cdot(2x-4)$$


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