Square Root Rule in Finding the Derivative

$\mathit{F}\mathit{i}\mathit{n}\mathit{d}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{D}\mathit{e}\mathit{r}\mathit{i}\mathit{v}\mathit{a}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{:} \mathit{y}\mathbf{=}\sqrt{\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{1}}$

$\mathit{}\mathit{S}\mathit{t}\mathit{e}\mathit{p}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathit{M}\mathit{u}\mathit{l}\mathit{t}\mathit{i}\mathit{p}\mathit{l}\mathit{y}\mathbf{ }\mathit{b}\mathit{o}\mathit{t}\mathit{h}\mathbf{ }\mathit{s}\mathit{i}\mathit{d}\mathit{e}\mathbf{ }\mathit{b}\mathit{y}\mathbf{ }\frac{\mathit{d}}{\mathit{d}\mathit{x}}$

$$\begin{gathered} \text{ \\ dydx=ddx3x2-5x+1 \\ } \end{gathered}$$

$\mathit{S}\mathit{t}\mathit{e}\mathit{p}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathit{A}\mathit{p}\mathit{p}\mathit{l}\mathit{y}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{S}\mathit{q}\mathit{u}\mathit{a}\mathit{r}\mathit{e}\mathbf{ }\mathit{R}\mathit{o}\mathit{o}\mathit{t}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{a}\mathbf{ }\mathit{F}\mathit{u}\mathit{n}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{R}\mathit{u}\mathit{l}\mathit{e} \mathit{i}\mathit{n}\mathbf{ }\frac{\mathbf{d}}{\mathbf{dx}}\sqrt{\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{1}}$

$$\begin{gathered} \text{ \\ ddxu=12uddx(u) \\ } \end{gathered}$$

$\mathbf{G}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{:}$

$$\begin{gathered} \text{ \\ u=3x2-5x+1} \end{gathered}$$

$$\begin{gathered} \text{ \\ dudx=6x-5 \\ } \end{gathered}$$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{3x^2-5x+1}}6\mathrm{x}-5$

$$\begin{gathered} \text{ \\ dydx=6x-523x2-5x+1 \\ } \end{gathered}$$

$\mathit{F}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathbf{:}\mathbf{ }\frac{\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{5}}{\mathbf{2}\sqrt{\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{1}}}$