\u00a0Recorrendo \u00e0 defini\u00e7\u00e3o de derivada de uma fun\u00e7\u00e3o num ponto, calcule a derivada de $f$ em $a$:<\/p>\n
\n$f:x \\to 2{x^2} – 3x$, em $a =\u00a0 – 1$<\/p>\n<\/blockquote>\n
$$\\begin{array}{*{20}{l}}
\n{f'( – 1)}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{f( – 1 + h) – f( – 1)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{2{{\\left( { – 1 + h} \\right)}^2} – 3\\left( { – 1 + h} \\right) + 5}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{2 – 4h + 2{h^2} + 3 – 3h + 5}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{h\\left( {2h – 7} \\right)}}{h}} \\\\
\n{}& = &{ – 7}
\n\\end{array}$$<\/p>\n<< Enunciado<\/a><\/span>R2 >><\/a><\/span><\/div><\/div>\n\n\nR2<\/b><\/span><\/p>\n\n$f:x \\to {x^3} – 1$, em $a = 0$ e em $a = 1$<\/p>\n<\/blockquote>\n
$$\\begin{array}{*{20}{l}}
\n{f'(0)}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{f(0 + h) – f(0)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{{{\\left( {0 + h} \\right)}^3} – 1 – ( – 1)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{{h^3}}}{h}} \\\\
\n{}& = &0
\n\\end{array}$$
\n$$\\begin{array}{*{20}{l}}
\n{f'(1)}& = &{\\mathop {\\lim }\\limits_{x \\to 1} \\frac{{f(x) – f(1)}}{{x – 1}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to 1} \\frac{{{x^3} – 1 – 0}}{{x – 1}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to 1} \\frac{{\\left( {x – 1} \\right)\\left( {{x^2} + x + 1} \\right)}}{{x – 1}}} \\\\
\n{}& = &3
\n\\end{array}$$<\/p>\n<< R1<\/a><\/span>R3 >><\/a><\/span><\/div><\/div>\n\n\nR3<\/b><\/span><\/p>\n\n$f:x \\to \\frac{1}{{{x^2}}}$, em $a =\u00a0 – 2$<\/p>\n<\/blockquote>\n
$$\\begin{array}{*{20}{l}}
\n{f'( – 2)}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 – 2} \\frac{{f(x) – f( – 2)}}{{x + 2}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 – 2} \\frac{{\\frac{1}{{{x^2}}} – \\frac{1}{4}}}{{x + 2}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 – 2} \\frac{{\\frac{{4 – {x^2}}}{{4{x^2}}}}}{{x – 1}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 – 2} \\frac{{4\\left( {1 + x} \\right)\\left( {1 – x} \\right)}}{{4{x^2}\\left( {x – 1} \\right)}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to\u00a0 – 2} \\frac{{ – \\left( {1 + x} \\right)}}{{{x^2}}}} \\\\
\n{}& = &{\\frac{1}{4}}
\n\\end{array}$$<\/p>\n<< R2<\/a><\/span>R4 >><\/a><\/span><\/div><\/div>\n\n\nR4<\/b><\/span><\/p>\n\n$f:x \\to \\frac{{3x + 2}}{{x – 5}}$, em $a = 4$<\/p>\n<\/blockquote>\n
$$\\begin{array}{*{20}{l}}
\n{f'(4)}& = &{\\mathop {\\lim }\\limits_{x \\to 4} \\frac{{f(x) – f(4)}}{{x – 4}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to 4} \\frac{{\\frac{{3x + 2}}{{x – 5}} – ( – 14)}}{{x – 4}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to 4} \\frac{{\\frac{{3x + 2 + 14x – 70}}{{x – 5}}}}{{x – 4}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{x \\to 4} \\frac{{17\\left( {x – 4} \\right)}}{{\\left( {x – 4} \\right)\\left( {x – 5} \\right)}}} \\\\
\n{}& = &{ – 17}
\n\\end{array}$$<\/p>\n<< R3<\/a><\/span>R5 >><\/a><\/span><\/div><\/div>\n\n\nR5<\/b><\/span><\/p>\n\n$f:x \\to \\sqrt x $, em $a = 4$<\/p>\n<\/blockquote>\n
$$\\begin{array}{*{20}{l}}
\n{f'(4)}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{f(4 + h) – f(4)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\sqrt {4 + h}\u00a0 – 2}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{\\left( {\\sqrt {4 + h}\u00a0 – 2} \\right)\\left( {\\sqrt {4 + h}\u00a0 + 2} \\right)}}{{h\\left( {\\sqrt {4 + h}\u00a0 + 2} \\right)}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{{4 + h – 4}}{{h\\left( {\\sqrt {4 + h}\u00a0 + 2} \\right)}}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to 0} \\frac{1}{{\\sqrt {4 + h}\u00a0 + 2}}} \\\\
\n{}& = &{\\frac{1}{4}}
\n\\end{array}$$<\/p>\n