Defina a derivada de cada uma das fun\u00e7\u00f5es:<\/p>\n
\n$f:x \\to {x^6} – 3{x^5} + 2{x^4} + x + 2$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {{x^6} – 3{x^5} + 2{x^4} + x + 2} \\right)}^\\prime }} \\\\
\n{}& = &{6{x^5} – 15{x^4} + 8{x^3} + 1}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to 6{x^5} – 15{x^4} + 8{x^3} + 1}
\n\\end{array}$$<\/p>\n<< Enunciado<\/a><\/span>R2 >><\/a><\/span><\/div><\/div>\n\n\nR2<\/b><\/span><\/p>\n\n$f:x \\to \\frac{1}{3}{x^4} – \\frac{1}{2}{x^3} – 3{x^2} + \\frac{1}{5}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{1}{3}{x^4} – \\frac{1}{2}{x^3} – 3{x^2} + \\frac{1}{5}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{4}{3}{x^3} – \\frac{3}{2}{x^2} – 6x}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{4}{3}{x^3} – \\frac{3}{2}{x^2} – 6x}
\n\\end{array}$$<\/p>\n<< R1<\/a><\/span>R3 >><\/a><\/span><\/div><\/div>\n\n\nR3<\/b><\/span><\/p>\n\n$f:x \\to \\pi {x^5} + \\frac{1}{2}{x^2} + \\sqrt 3 $<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\pi {x^5} + \\frac{1}{2}{x^2} + \\sqrt 3 } \\right)}^\\prime }} \\\\
\n{}& = &{5\\pi {x^4} + x}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to 5\\pi {x^4} + x}
\n\\end{array}$$<\/p>\n<< R2<\/a><\/span>R4 >><\/a><\/span><\/div><\/div>\n\n\nR4<\/b><\/span><\/p>\n\n$f:x \\to \\frac{2}{{3{x^2} – 3}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:3{x^2} – 3 \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – 1,1} \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{2}{{3{x^2} – 3}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{ – 2 \\times {{\\left( {3{x^2} – 3} \\right)}^\\prime }}}{{{{\\left( {3{x^2} – 3} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{ – 12x}}{{9{{\\left( {{x^2} – 1} \\right)}^2}}}} \\\\
\n{}& = &{ – \\frac{{4x}}{{3{{\\left( {{x^2} – 1} \\right)}^2}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R}\\backslash \\left\\{ { – 1,1} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to\u00a0 – \\frac{{4x}}{{3{{\\left( {{x^2} – 1} \\right)}^2}}}}
\n\\end{array}$$<\/p>\n<< R3<\/a><\/span>R5 >><\/a><\/span><\/div><\/div>\n\n\nR5<\/b><\/span><\/p>\n\n$f:x \\to \\frac{{{x^2} + 1}}{{3{x^2} + x + 1}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:3{x^2} + x + 1 \\ne 0} \\right\\} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{{{x^2} + 1}}{{3{x^2} + x + 1}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\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}}}} \\\\
\n{}& = &{\\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}}}} \\\\
\n{}& = &{\\frac{{6{x^3} + 2{x^2} + 2x – 6{x^3} – 6x – {x^2} – 1}}{{{{\\left( {3{x^2} + x + 1} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{{x^2} – 4x – 1}}{{{{\\left( {3{x^2} + x + 1} \\right)}^2}}}} \\\\
\n{}&{}&{}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{{x^2} – 4x – 1}}{{{{\\left( {3{x^2} + x + 1} \\right)}^2}}}}
\n\\end{array}$$<\/p>\n<< R4<\/a><\/span>R6 >><\/a><\/span><\/div><\/div>\n\n\nR6<\/b><\/span><\/p>\n\n$f:x \\to {\\left( {2x + 1} \\right)^3}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {{{\\left( {2x + 1} \\right)}^3}} \\right)}^\\prime }} \\\\
\n{}& = &{3{{\\left( {2x + 1} \\right)}^2}{{\\left( {2x + 1} \\right)}^\\prime }} \\\\
\n{}& = &{6{{\\left( {2x + 1} \\right)}^2}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to 6{{\\left( {2x + 1} \\right)}^2}}
\n\\end{array}$$<\/p>\n<< R5<\/a><\/span>R7 >><\/a><\/span><\/div><\/div>\n\n\nR7<\/b><\/span><\/p>\n\n$f:x \\to 1 – \\sqrt x $<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:x \\geqslant 0} \\right\\} = \\mathbb{R}_0^ + $$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {1 – {x^{\\frac{1}{2}}}} \\right)}^\\prime }} \\\\
\n{}& = &{ – \\left( {\\frac{1}{2}{x^{ – \\frac{1}{2}}} \\times 1} \\right)} \\\\
\n{}& = &{ – \\frac{1}{{2\\sqrt x }}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{{\\mathbb{R}^ + } \\to \\mathbb{R}} \\\\
\n{}&{x \\to\u00a0 – \\frac{1}{{2\\sqrt x }}}
\n\\end{array}$$<\/p>\n<< R6<\/a><\/span>R8 >><\/a><\/span><\/div><\/div>\n\n\nR8<\/b><\/span><\/p>\n\n$f:x \\to\u00a0 – \\frac{1}{{3{x^2}}} + \\frac{1}{x}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:3{x^2} \\ne 0 \\wedge x \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( { – \\frac{1}{{3{x^2}}} + \\frac{1}{x}} \\right)}^\\prime }} \\\\
\n{}& = &{ – \\frac{{ – 6x}}{{{{\\left( {3{x^2}} \\right)}^2}}} + \\frac{{ – 1}}{{{x^2}}}} \\\\
\n{}& = &{\\frac{2}{{3{x^3}}} – \\frac{1}{{{x^2}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{2}{{3{x^3}}} – \\frac{1}{{{x^2}}}}
\n\\end{array}$$<\/p>\n<< R7<\/a><\/span>R9 >><\/a><\/span><\/div><\/div>\n\n\nR9<\/b><\/span><\/p>\n\n$f:x \\to {\\left( {\\frac{{x + 1}}{{2x – 3}}} \\right)^2}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:2x – 3 \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ {\\frac{3}{2}} \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {{{\\left( {\\frac{{x + 1}}{{2x – 3}}} \\right)}^2}} \\right)}^\\prime }} \\\\
\n{}& = &{2 \\times \\frac{{x + 1}}{{2x – 3}} \\times {{\\left( {\\frac{{x + 1}}{{2x – 3}}} \\right)}^\\prime }} \\\\
\n{}& = &{2 \\times \\frac{{x + 1}}{{2x – 3}} \\times \\frac{{\\left( {2x – 3} \\right) – 2\\left( {x + 1} \\right)}}{{{{\\left( {2x – 3} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{ – 10x – 10}}{{{{\\left( {2x – 3} \\right)}^3}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R}\\backslash \\left\\{ {\\frac{3}{2}} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{ – 10x – 10}}{{{{\\left( {2x – 3} \\right)}^3}}}}
\n\\end{array}$$<\/p>\n<< R8<\/a><\/span>R10 >><\/a><\/span><\/div><\/div>\n\n\nR10<\/b><\/span><\/p>\n\n$f:x \\to \\frac{1}{{{{\\left( {3x + 1} \\right)}^2}}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:3x + 1 \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – \\frac{1}{3}} \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{1}{{{{\\left( {3x + 1} \\right)}^2}}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{0 – 2\\left( {3x + 1} \\right) \\times 3 \\times 1}}{{{{\\left( {3x + 1} \\right)}^4}}}} \\\\
\n{}& = &{ – \\frac{6}{{{{\\left( {3x + 1} \\right)}^3}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R}\\backslash \\left\\{ { – \\frac{1}{3}} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to\u00a0 – \\frac{6}{{{{\\left( {3x + 1} \\right)}^3}}}}
\n\\end{array}$$<\/p>\n<< R9<\/a><\/span>R11 >><\/a><\/span><\/div><\/div>\n\n\nR11<\/b><\/span><\/p>\n\n$f:x \\to \\left( {3x + 1} \\right){\\left( {2x – 1} \\right)^2}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\left( {3x + 1} \\right){{\\left( {2x – 1} \\right)}^2}} \\right)}^\\prime }} \\\\
\n{}& = &{3 \\times {{\\left( {2x – 1} \\right)}^2} + 2\\left( {2x – 1} \\right) \\times 2 \\times \\left( {3x + 1} \\right)} \\\\
\n{}& = &{\\left( {2x – 1} \\right)\\left[ {3\\left( {2x – 1} \\right) + 4\\left( {3x + 1} \\right)} \\right]} \\\\
\n{}& = &{\\left( {2x – 1} \\right)\\left( {18x + 1} \\right)}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\left( {2x – 1} \\right)\\left( {18x + 1} \\right)}
\n\\end{array}$$<\/p>\n<< R10<\/a><\/span>R12 >><\/a><\/span><\/div><\/div>\n\n\nR12<\/b><\/span><\/p>\n\n$f:x \\to x\\left( {x + 3} \\right)\\left( {2x – 1} \\right)$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {x\\left( {x + 3} \\right)\\left( {2x – 1} \\right)} \\right)}^\\prime }} \\\\
\n{}& = &{1 \\times \\left( {x + 3} \\right)\\left( {2x – 1} \\right) + 1 \\times x\\left( {2x – 1} \\right) + 2 \\times x\\left( {x + 3} \\right)} \\\\
\n{}& = &{2{x^2} + 5x – 3 + 2{x^2} – x + 2{x^2} + 6x} \\\\
\n{}& = &{6{x^2} + 10x – 3}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to 6{x^2} + 10x – 3}
\n\\end{array}$$<\/p>\n<< R11<\/a><\/span>R13 >><\/a><\/span><\/div><\/div>\n\n\nR13<\/b><\/span><\/p>\n\n$f:x \\to \\frac{{{x^2}}}{4} – \\frac{x}{5}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{1}{4}{x^2} – \\frac{1}{5}x} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{1}{4} \\times 2x – \\frac{1}{5} \\times 1} \\\\
\n{}& = &{\\frac{x}{2} – \\frac{1}{5}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{x}{2} – \\frac{1}{5}}
\n\\end{array}$$<\/p>\n<< R12<\/a><\/span>R14 >><\/a><\/span><\/div><\/div>\n\n\nR14<\/b><\/span><\/p>\n\n$f:x \\to 3x – 1 – \\frac{4}{{2x + 1}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:2x + 1 \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – \\frac{1}{2}} \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {3x – 1 – \\frac{4}{{2x + 1}}} \\right)}^\\prime }} \\\\
\n{}& = &{3 – \\frac{{0 – 2 \\times 4}}{{{{\\left( {2x + 1} \\right)}^2}}}} \\\\
\n{}& = &{3 + \\frac{8}{{{{\\left( {2x + 1} \\right)}^2}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R}\\backslash \\left\\{ { – \\frac{1}{2}} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to 3 + \\frac{8}{{{{\\left( {2x + 1} \\right)}^2}}}}
\n\\end{array}$$<\/p>\n