Determine uma express\u00e3o anal\u00edtica da derivada de cada uma das fun\u00e7\u00f5es:<\/p>\n
\n$f:x \\to {e^{ – 4x}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(x)}& = &{\\left( {{e^{ – 4x}}} \\right)’} \\\\
\n{}& = &{\\left( { – 4x} \\right)'{e^{ – 4x}}} \\\\
\n{}& = &{ – 4{e^{ – 4x}}}
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to\u00a0 – 4{e^{ – 4x}}}
\n\\end{array}$$<\/p>\n<< Enunciado<\/a><\/span>R2 >><\/a><\/span><\/div><\/div>\n\n\nR2<\/b><\/span><\/p>\n\n$f:x \\to {e^{\\sqrt {2 + x} }}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:2 + x \\geqslant 0} \\right\\} = \\left[ { – 2, + \\infty } \\right[$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(x)}& = &{\\left( {{e^{\\sqrt {2 + x} }}} \\right)’} \\\\
\n{}& = &{\\left( {\\sqrt {2 + x} } \\right)'{e^{\\sqrt {2 + x} }}} \\\\
\n{}& = &{\\frac{1}{2}{{\\left( {2 + x} \\right)}^{ – \\tfrac{1}{2}}} \\times {e^{\\sqrt {2 + x} }}} \\\\
\n{}& = &{\\frac{{{e^{\\sqrt {2 + x} }}}}{{2\\sqrt {2 + x} }}}
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\left] { – 2, + \\infty } \\right[ \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{{e^{\\sqrt {2 + x} }}}}{{2\\sqrt {2 + x} }}}
\n\\end{array}$$<\/p>\n<< R1<\/a><\/span>R3 >><\/a><\/span><\/div><\/div>\n\n\nR3<\/b><\/span><\/p>\n\n$f:x \\to {e^x}\\left( {{x^2} + 2x + 3} \\right)$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(x)}& = &{\\left( {{e^x}\\left( {{x^2} + 2x + 3} \\right)} \\right)’} \\\\
\n{}& = &{{e^x}\\left( {{x^2} + 2x + 3} \\right) + {e^x}\\left( {2x + 2} \\right)} \\\\
\n{}& = &{{e^x}\\left( {{x^2} + 4x + 5} \\right)}
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to {e^x}\\left( {{x^2} + 4x + 5} \\right)}
\n\\end{array}$$<\/p>\n<< R2<\/a><\/span>R4 >><\/a><\/span><\/div><\/div>\n\n\nR4<\/b><\/span><\/p>\n\n$f:x \\to {e^{\\frac{1}{x}}} + {e^{ – \\frac{1}{x}}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {{e^{\\frac{1}{x}}} + {e^{ – \\frac{1}{x}}}} \\right)}^\\prime }} \\\\
\n{}& = &{ – \\frac{1}{{{x^2}}}{e^{\\frac{1}{x}}} + \\frac{1}{{{x^2}}}{e^{ – \\frac{1}{x}}}} \\\\
\n{}& = &{\\frac{1}{{{x^2}}}\\left( {{e^{ – \\frac{1}{x}}} – {e^{\\frac{1}{x}}}} \\right)}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{1}{{{x^2}}}\\left( {{e^{ – \\frac{1}{x}}} – {e^{\\frac{1}{x}}}} \\right)}
\n\\end{array}$$<\/p>\n<< R3<\/a><\/span>R5 >><\/a><\/span><\/div><\/div>\n\n\nR5<\/b><\/span><\/p>\n\n$f:x \\to \\frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{\\left( {{e^x} + {e^{ – x}}} \\right)\\left( {{e^x} + {e^{ – x}}} \\right) – \\left( {{e^x} – {e^{ – x}}} \\right)\\left( {{e^x} – {e^{ – x}}} \\right)}}{{{{\\left( {{e^x} + {e^{ – x}}} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{{e^{2x}} + 1 + 1 + {e^{ – 2x}} – {e^{2x}} + 1 + 1 – {e^{ – 2x}}}}{{{{\\left( {{e^x} + {e^{ – x}}} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{4}{{{{\\left( {{e^x} + {e^{ – x}}} \\right)}^2}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{4}{{{{\\left( {{e^x} + {e^{ – x}}} \\right)}^2}}}}
\n\\end{array}$$<\/p>\n<< R4<\/a><\/span>R6 >><\/a><\/span><\/div><\/div>\n\n\nR6<\/b><\/span><\/p>\n\n$f:x \\to \\frac{x}{{{e^x}}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\mathbb{R}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{x}{{{e^x}}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{1 \\times {e^x} – {e^x} \\times x}}{{{{\\left( {{e^x}} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{\\left( {1 – x} \\right){e^x}}}{{{e^{2x}}}}} \\\\
\n{}& = &{\\frac{{1 – x}}{{{e^x}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{1 – x}}{{{e^x}}}}
\n\\end{array}$$<\/p>\n<< R5<\/a><\/span>R7 >><\/a><\/span><\/div><\/div>\n\n\nR7<\/b><\/span><\/p>\n\n$f:x \\to \\ln \\left( {3x – 5} \\right)$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:3x – 5 > 0} \\right\\} = \\left] {\\frac{5}{3}, + \\infty } \\right[$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(x)}& = &{\\left( {\\ln \\left( {3x – 5} \\right)} \\right)’} \\\\
\n{}& = &{\\frac{{\\left( {3x – 5} \\right)’}}{{3x – 5}}} \\\\
\n{}& = &{\\frac{3}{{3x – 5}}}
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\left] {\\frac{5}{3}, + \\infty } \\right[ \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{3}{{3x – 5}}}
\n\\end{array}$$<\/p>\n<< R6<\/a><\/span>R8 >><\/a><\/span><\/div><\/div>\n\n\nR8<\/b><\/span><\/p>\n\n$f:x \\leftarrow x\\ln x + {e^3}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:x > 0} \\right\\} = {\\mathbb{R}^ + }$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(x)}& = &{\\left( {x\\ln x + {e^3}} \\right)’} \\\\
\n{}& = &{1 \\times \\ln x + x \\times \\frac{{x’}}{x}} \\\\
\n{}& = &{\\ln x + 1}
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{{\\mathbb{R}^ + } \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\ln x + 1}
\n\\end{array}$$<\/p>\n<< R7<\/a><\/span>R9 >><\/a><\/span><\/div><\/div>\n\n\nR9<\/b><\/span><\/p>\n\n$f:x \\to \\ln \\left( {\\ln x} \\right)$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:x > 0 \\wedge \\ln x > 0} \\right\\} = \\left] {1, + \\infty } \\right[$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(x)}& = &{\\left( {\\ln \\left( {\\ln x} \\right)} \\right)’} \\\\
\n{}& = &{\\frac{{\\left( {\\ln x} \\right)’}}{{\\ln x}}} \\\\
\n{}& = &{\\frac{{\\frac{1}{x}}}{{\\ln x}}} \\\\
\n{}& = &{\\frac{1}{{x\\ln x}}}
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\left] {1, + \\infty } \\right[ \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{1}{{x\\ln x}}}
\n\\end{array}$$<\/p>\n<< R8<\/a><\/span>R10 >><\/a><\/span><\/div><\/div>\n\n\nR10<\/b><\/span><\/p>\n\n$f:x \\to \\frac{{{x^2}}}{{\\ln x}}$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:x > 0 \\wedge \\ln x \\ne 0} \\right\\} = {\\mathbb{R}^ + }\\backslash \\left\\{ 1 \\right\\}$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {\\frac{{{x^2}}}{{\\ln x}}} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{2x \\times \\ln x – \\frac{{x’}}{x} \\times {x^2}}}{{{{\\left( {\\ln x} \\right)}^2}}}} \\\\
\n{}& = &{\\frac{{2x\\ln x – x}}{{{{\\left( {\\ln x} \\right)}^2}}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{{\\mathbb{R}^ + }\\backslash \\left\\{ 1 \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{2x\\ln x – x}}{{{{\\left( {\\ln x} \\right)}^2}}}}
\n\\end{array}$$<\/p>\n<< R9<\/a><\/span>R11 >><\/a><\/span><\/div><\/div>\n\n\nR11<\/b><\/span><\/p>\n\n$f:x \\to {\\log _2}\\left( {{x^2} – 4x} \\right)$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:{x^2} – 4x > 0} \\right\\} = \\left] { – \\infty ,0} \\right[ \\cup \\left] {4, + \\infty } \\right[$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {{{\\log }_2}\\left( {{x^2} – 4x} \\right)} \\right)}^\\prime }} \\\\
\n{}& = &{\\frac{{{{\\left( {{x^2} – 4x} \\right)}^\\prime }}}{{\\left( {{x^2} – 4x} \\right)\\ln 2}}} \\\\
\n{}& = &{\\frac{{2x – 4}}{{\\left( {{x^2} – 4x} \\right)\\ln 2}}}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{\\left] { – \\infty ,0} \\right[ \\cup \\left] {4, + \\infty } \\right[ \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{2x – 4}}{{\\left( {{x^2} – 4x} \\right)\\ln 2}}}
\n\\end{array}$$<\/p>\n<< R10<\/a><\/span>R12 >><\/a><\/span><\/div><\/div>\n\n\nR12<\/b><\/span><\/p>\n\n$f:x \\to {e^{3x}}\\ln x$<\/p>\n<\/blockquote>\n
$${D_f} = \\left\\{ {x \\in \\mathbb{R}:x > 0} \\right\\} = {\\mathbb{R}^ + }$$
\n\\[\\begin{array}{*{20}{l}}
\n{f’\\left( x \\right)}& = &{{{\\left( {{e^{3x}}\\ln x} \\right)}^\\prime }} \\\\
\n{}& = &{3{e^{3x}} \\times \\ln x + {e^{3x}} \\times \\frac{1}{x}} \\\\
\n{}& = &{{e^{3x}}\\left( {3\\ln x + \\frac{1}{x}} \\right)}
\n\\end{array}\\]
\n$$\\begin{array}{*{20}{l}}
\n{f’:}&{{\\mathbb{R}^ + } \\to \\mathbb{R}} \\\\
\n{}&{x \\to {e^{3x}}\\left( {3\\ln x + \\frac{1}{x}} \\right)}
\n\\end{array}$$<\/p>\n