Find the second derivative of the root function

Find the second derivative of $y=\sqrt{3x^2-2x+1}$

By indices

$y=(3x^2-2x+1)^\frac{1}{2}$

We use chain rule to evaluate $y$. 

Formula $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

Let $u=3x^2-2x+1, \frac{du}{dx}=6x-2$

And $u^\frac{1}{2}, \frac{dy}{du}=\frac{1}{2}u^\frac{-1}{2}$. 

Substitute $y,u$ and their derivatives into the chain rule formula.

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\frac{1}{2}u^\frac{-1}{2}\times{6x}=3xu^\frac{-1}{2}$

$\frac{dy}{dx}=3x(3x^2-2x+1)^\frac{-1}{2}$.

That is all for the first derivative.

To find the second derivative $\frac{d^2y}{dx^2}$ of $\frac{dy}{dx}=3x(3x^2-2x+1)^\frac{-1}{2}$, we use product rule.

Formula for product rule $\frac{d^2y}{dx^2}=V\frac{du}{dx}+U\frac{dv}{dx}$. 

Let $V=3x, \frac{dv}{dx}=3, U=(3x^2-2x+1)^\frac{-1}{2}, \frac{du}{dx}=\frac{-1}{2}(3x^2-2x+1)^\frac{-3}{2}$. 

$\frac{d^2y}{dx^2}=3x(\frac{-1}{2}(3x^2-2x+1)^\frac{-3}{2})+(3x^2-2x+1)^\frac{-1}{2}.3$

 $=\frac{-3x}{2}(3x^2-2x+1)^\frac{-3}{2}+3(3x^2-2x+1)^\frac{-1}{2}$

Factor out common factors if you like.