\begin{enumerate}
\item $f(x)=2x\sqrt{x}$. On pose :

\begin{multicols}{2}
$u(x)=2x$

$u'(x)=2$

\columnbreak

$v(x)=\sqrt{x}$

$v'(x)=\dfrac{1}{2\sqrt{x}}$
\end{multicols}

On a alors :

\begin{align*}
f'(x) & = u'(x)v(x)+u(x)v'(x)\\
& = 2\sqrt{x}+2x\times\dfrac{1}{2\sqrt{x}}\\
& = 2\sqrt{x}+\dfrac{x}{\sqrt x}\\
& = 2\sqrt{x}+\dfrac{\cancel{\sqrt{x}}\times\sqrt{x}}{\cancel{\sqrt x}}\\
& = 2\sqrt{x}+\sqrt{x}\\
\Aboxed{f'(x) & = 3\sqrt{x}}
\end{align*}


\item $f(x)=\dfrac{3x-1}{4x+5}$. On pose :

\begin{multicols}{2}
$u(x)=3x-1$

$u'(x)=3$

\columnbreak

$v(x)=4x+5$

$v'(x)=4$
\end{multicols}

On a alors :

\begin{align*}
f'(x) & = \dfrac{u'(x)v(x)-u(x)v'(x)}{\big[v(x)\big]^2}\\
&= \dfrac{3(4x+5)-4(3x-1)}{(4x+5)^2}\\
& = \dfrac{12x+15-12x+4}{(4x+5)^2}\\
\Aboxed{f'(x) & = \dfrac{19}{(4x+5)^2}}
\end{align*}
\newpage
\item $f(x)=\dfrac{\sqrt{x}-x}{x^2+1}$. On pose :

\begin{multicols}{2}
$u(x)=\sqrt{x}-x$

$u'(x)=\dfrac{1}{2\sqrt{x}}-1$

\columnbreak

$v(x)=x^2+1$

$v'(x)=2x$
\end{multicols}

On a alors :
\begin{align*}
f'(x) & = 
\dfrac{u'(x)v(x)-u(x)v'(x)}{\big[v(x)\big]^2}\\
& = 
\dfrac{
\left(\dfrac{1}{2\sqrt{x}}-1\right)\times(x^2+1)-\left(\sqrt{x}-x\right)\times2x}{\big(x^2+1\big)^2}\\
& = \dfrac{\dfrac{x^2+1}{2\sqrt{x}}-x^2-1-2x\sqrt{x}+2x^2}{\big(x^2+1\big)^2}\\
& = \dfrac{\dfrac{x^2+1}{2\sqrt{x}}+x^2-2x\sqrt{x}-1}{\big(x^2+1\big)^2}\\
& = \dfrac{\dfrac{x^2+1}{2\sqrt{x}}+\dfrac{2x^2\sqrt{x}}{2\sqrt{x}}-\dfrac{4x^2}{2\sqrt{x}}-\dfrac{2\sqrt{x}}{2\sqrt{x}}}{\big(x^2+1\big)^2}\\
& = \dfrac{\dfrac{2x^2\sqrt{x}-2\sqrt{x}-3x^2+1}{2\sqrt{x}}}{\big(x^2+1\big)^2}\\
\Aboxed{f'(x) & = \dfrac{2x^2\sqrt{x}-2\sqrt{x}-3x^2+1}{2\sqrt{x}\big(x^2+1\big)^2}}	
\end{align*}

\item $f(x)=\dfrac{x^2-3x+1}{x-3}$. On pose :

\begin{multicols}{2}
$u(x)=x^2-3x+1$

$u'(x)=2x-3$

\columnbreak

$v(x)=x-3$

$v'(x)=1$
\end{multicols}

On a donc :
\begin{align*}
f'(x) & = \dfrac{u'(x)v(x)-u(x)v'(x)}{\big[v(x)\big]^2}\\
& = \dfrac{(2x-3)(x-3)-(x^2-3x+1)\times1}{(x-3)^2}\\
& = \dfrac{2x^2-6x-3x+9-x^2+3x-1}{(x-3)^2}\\
\Aboxed{f'(x) & = \dfrac{x^2-6x+8}{(x-3)^2}}
\end{align*}

\item $f(x)=\dfrac{\sqrt x}{x+\dfrac{1}{x}}$.

On commence par simplifier l'écriture de $f(x)$ :

\begin{align*}
f(x) & = \dfrac{\sqrt{x}}{\frac{x^2+1}{x}}\\
& = \dfrac{x\sqrt{x}}{x^2+1}.
\end{align*}

On pose alors :

\begin{multicols}{2}
$u(x)=x\sqrt{x}$

$\begin{aligned}[t]
u'(x) & =1\times\sqrt{x}+x\times\dfrac{1}{2\sqrt{x}}\\
& = \sqrt{x}+\dfrac{\cancel{\sqrt{x}}\times\sqrt{x}}{2\cancel{\sqrt x}}\\
& = \sqrt{x}+\dfrac{1}{2}\sqrt{x}\\
&=  \dfrac{3}{2}\sqrt{x}
\end{aligned}$

\columnbreak

$v(x)=x^2+1$

$v'(x)=2x$
\end{multicols}

D'où :
\begin{align*}
f'(x) & = \dfrac{u'(x)v(x)-u(x)v'(x)}{\big[v(x)\big]^2}\\
& = \dfrac{\frac{3}{2}\sqrt x(x^2+1)-x\sqrt x \times 2x}{(x^2+1)^2}\\
& = \dfrac{\frac32x^2\sqrt x+\frac32\sqrt x-2x^2\sqrt x}{(x^2+1)^2}\\
& = \dfrac{\frac32\sqrt x-\frac12x^2\sqrt x}{(x^2+1)^2}\\
\Aboxed{f'(x) & = \dfrac{\frac12\sqrt x\big(3-x^2\big)}{(x^2+1)^2}}
\end{align*}
\end{enumerate}