Defina a derivada de cada uma das funções

$f:x \to {x^6} – 3{x^5} + 2{x^4} + x + 2$

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {{x^6} – 3{x^5} + 2{x^4} + x + 2} \right)}^\prime }} \\
{}& = &{6{x^5} – 15{x^4} + 8{x^3} + 1}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to 6{x^5} – 15{x^4} + 8{x^3} + 1}
\end{array}$$

$f:x \to \frac{1}{3}{x^4} – \frac{1}{2}{x^3} – 3{x^2} + \frac{1}{5}$

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\frac{1}{3}{x^4} – \frac{1}{2}{x^3} – 3{x^2} + \frac{1}{5}} \right)}^\prime }} \\
{}& = &{\frac{4}{3}{x^3} – \frac{3}{2}{x^2} – 6x}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to \frac{4}{3}{x^3} – \frac{3}{2}{x^2} – 6x}
\end{array}$$

$f:x \to \pi {x^5} + \frac{1}{2}{x^2} + \sqrt 3 $

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\pi {x^5} + \frac{1}{2}{x^2} + \sqrt 3 } \right)}^\prime }} \\
{}& = &{5\pi {x^4} + x}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to 5\pi {x^4} + x}
\end{array}$$

$f:x \to \frac{2}{{3{x^2} – 3}}$

$${D_f} = \left\{ {x \in \mathbb{R}:3{x^2} – 3 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – 1,1} \right\}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\frac{2}{{3{x^2} – 3}}} \right)}^\prime }} \\
{}& = &{\frac{{ – 2 \times {{\left( {3{x^2} – 3} \right)}^\prime }}}{{{{\left( {3{x^2} – 3} \right)}^2}}}} \\
{}& = &{\frac{{ – 12x}}{{9{{\left( {{x^2} – 1} \right)}^2}}}} \\
{}& = &{ – \frac{{4x}}{{3{{\left( {{x^2} – 1} \right)}^2}}}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R}\backslash \left\{ { – 1,1} \right\} \to \mathbb{R}} \\
{}&{x \to  – \frac{{4x}}{{3{{\left( {{x^2} – 1} \right)}^2}}}}
\end{array}$$

$f:x \to \frac{{{x^2} + 1}}{{3{x^2} + x + 1}}$

$${D_f} = \left\{ {x \in \mathbb{R}:3{x^2} + x + 1 \ne 0} \right\} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\frac{{{x^2} + 1}}{{3{x^2} + x + 1}}} \right)}^\prime }} \\
{}& = &{\frac{{{{\left( {{x^2} + 1} \right)}^\prime }\left( {3{x^2} + x + 1} \right) – {{\left( {3{x^2} + x + 1} \right)}^\prime }\left( {{x^2} + 1} \right)}}{{{{\left( {3{x^2} + x + 1} \right)}^2}}}} \\
{}& = &{\frac{{2x\left( {3{x^2} + x + 1} \right) – \left( {6x + 1} \right)\left( {{x^2} + 1} \right)}}{{{{\left( {3{x^2} + x + 1} \right)}^2}}}} \\
{}& = &{\frac{{6{x^3} + 2{x^2} + 2x – 6{x^3} – 6x – {x^2} – 1}}{{{{\left( {3{x^2} + x + 1} \right)}^2}}}} \\
{}& = &{\frac{{{x^2} – 4x – 1}}{{{{\left( {3{x^2} + x + 1} \right)}^2}}}} \\
{}&{}&{}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to \frac{{{x^2} – 4x – 1}}{{{{\left( {3{x^2} + x + 1} \right)}^2}}}}
\end{array}$$

$f:x \to {\left( {2x + 1} \right)^3}$

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {{{\left( {2x + 1} \right)}^3}} \right)}^\prime }} \\
{}& = &{3{{\left( {2x + 1} \right)}^2}{{\left( {2x + 1} \right)}^\prime }} \\
{}& = &{6{{\left( {2x + 1} \right)}^2}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to 6{{\left( {2x + 1} \right)}^2}}
\end{array}$$

$f:x \to 1 – \sqrt x $

$${D_f} = \left\{ {x \in \mathbb{R}:x \geqslant 0} \right\} = \mathbb{R}_0^ + $$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {1 – {x^{\frac{1}{2}}}} \right)}^\prime }} \\
{}& = &{ – \left( {\frac{1}{2}{x^{ – \frac{1}{2}}} \times 1} \right)} \\
{}& = &{ – \frac{1}{{2\sqrt x }}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{{\mathbb{R}^ + } \to \mathbb{R}} \\
{}&{x \to  – \frac{1}{{2\sqrt x }}}
\end{array}$$

$f:x \to  – \frac{1}{{3{x^2}}} + \frac{1}{x}$

$${D_f} = \left\{ {x \in \mathbb{R}:3{x^2} \ne 0 \wedge x \ne 0} \right\} = \mathbb{R}\backslash \left\{ 0 \right\}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( { – \frac{1}{{3{x^2}}} + \frac{1}{x}} \right)}^\prime }} \\
{}& = &{ – \frac{{ – 6x}}{{{{\left( {3{x^2}} \right)}^2}}} + \frac{{ – 1}}{{{x^2}}}} \\
{}& = &{\frac{2}{{3{x^3}}} – \frac{1}{{{x^2}}}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R}\backslash \left\{ 0 \right\} \to \mathbb{R}} \\
{}&{x \to \frac{2}{{3{x^3}}} – \frac{1}{{{x^2}}}}
\end{array}$$

$f:x \to {\left( {\frac{{x + 1}}{{2x – 3}}} \right)^2}$

$${D_f} = \left\{ {x \in \mathbb{R}:2x – 3 \ne 0} \right\} = \mathbb{R}\backslash \left\{ {\frac{3}{2}} \right\}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {{{\left( {\frac{{x + 1}}{{2x – 3}}} \right)}^2}} \right)}^\prime }} \\
{}& = &{2 \times \frac{{x + 1}}{{2x – 3}} \times {{\left( {\frac{{x + 1}}{{2x – 3}}} \right)}^\prime }} \\
{}& = &{2 \times \frac{{x + 1}}{{2x – 3}} \times \frac{{\left( {2x – 3} \right) – 2\left( {x + 1} \right)}}{{{{\left( {2x – 3} \right)}^2}}}} \\
{}& = &{\frac{{ – 10x – 10}}{{{{\left( {2x – 3} \right)}^3}}}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R}\backslash \left\{ {\frac{3}{2}} \right\} \to \mathbb{R}} \\
{}&{x \to \frac{{ – 10x – 10}}{{{{\left( {2x – 3} \right)}^3}}}}
\end{array}$$

$f:x \to \frac{1}{{{{\left( {3x + 1} \right)}^2}}}$

$${D_f} = \left\{ {x \in \mathbb{R}:3x + 1 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – \frac{1}{3}} \right\}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\frac{1}{{{{\left( {3x + 1} \right)}^2}}}} \right)}^\prime }} \\
{}& = &{\frac{{0 – 2\left( {3x + 1} \right) \times 3 \times 1}}{{{{\left( {3x + 1} \right)}^4}}}} \\
{}& = &{ – \frac{6}{{{{\left( {3x + 1} \right)}^3}}}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R}\backslash \left\{ { – \frac{1}{3}} \right\} \to \mathbb{R}} \\
{}&{x \to  – \frac{6}{{{{\left( {3x + 1} \right)}^3}}}}
\end{array}$$

$f:x \to \left( {3x + 1} \right){\left( {2x – 1} \right)^2}$

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\left( {3x + 1} \right){{\left( {2x – 1} \right)}^2}} \right)}^\prime }} \\
{}& = &{3 \times {{\left( {2x – 1} \right)}^2} + 2\left( {2x – 1} \right) \times 2 \times \left( {3x + 1} \right)} \\
{}& = &{\left( {2x – 1} \right)\left[ {3\left( {2x – 1} \right) + 4\left( {3x + 1} \right)} \right]} \\
{}& = &{\left( {2x – 1} \right)\left( {18x + 1} \right)}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to \left( {2x – 1} \right)\left( {18x + 1} \right)}
\end{array}$$

$f:x \to x\left( {x + 3} \right)\left( {2x – 1} \right)$

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {x\left( {x + 3} \right)\left( {2x – 1} \right)} \right)}^\prime }} \\
{}& = &{1 \times \left( {x + 3} \right)\left( {2x – 1} \right) + 1 \times x\left( {2x – 1} \right) + 2 \times x\left( {x + 3} \right)} \\
{}& = &{2{x^2} + 5x – 3 + 2{x^2} – x + 2{x^2} + 6x} \\
{}& = &{6{x^2} + 10x – 3}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to 6{x^2} + 10x – 3}
\end{array}$$

$f:x \to \frac{{{x^2}}}{4} – \frac{x}{5}$

$${D_f} = \mathbb{R}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {\frac{1}{4}{x^2} – \frac{1}{5}x} \right)}^\prime }} \\
{}& = &{\frac{1}{4} \times 2x – \frac{1}{5} \times 1} \\
{}& = &{\frac{x}{2} – \frac{1}{5}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R} \to \mathbb{R}} \\
{}&{x \to \frac{x}{2} – \frac{1}{5}}
\end{array}$$

$f:x \to 3x – 1 – \frac{4}{{2x + 1}}$

$${D_f} = \left\{ {x \in \mathbb{R}:2x + 1 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – \frac{1}{2}} \right\}$$
\[\begin{array}{*{20}{l}}
{f’\left( x \right)}& = &{{{\left( {3x – 1 – \frac{4}{{2x + 1}}} \right)}^\prime }} \\
{}& = &{3 – \frac{{0 – 2 \times 4}}{{{{\left( {2x + 1} \right)}^2}}}} \\
{}& = &{3 + \frac{8}{{{{\left( {2x + 1} \right)}^2}}}}
\end{array}\]
$$\begin{array}{*{20}{l}}
{f’:}&{\mathbb{R}\backslash \left\{ { – \frac{1}{2}} \right\} \to \mathbb{R}} \\
{}&{x \to 3 + \frac{8}{{{{\left( {2x + 1} \right)}^2}}}}
\end{array}$$