Recorrendo à definição de derivada

Cálculo diferencial: Infinito 12 A - Parte 2 Pág. 211 Ex. 37

by AMMA · Published 27 de Março de 2012 · Updated 30 de Dezembro de 2021

Enunciado

Recorrendo à definição de derivada de uma função num ponto, calcule a derivada de $f$ em $a$:

$f:x \to 2{x^2} – 3x$, em $a = – 1$;
$f:x \to {x^3} – 1$, em $a = 0$ e em $a = 1$;
$f:x \to \frac{1}{{{x^2}}}$, em $a = – 2$;
$f:x \to \frac{{3x + 2}}{{x – 5}}$, em $a = 4$;
$f:x \to \sqrt x $, em $a = 4$.

$f:x \to 2{x^2} – 3x$, em $a = – 1$

$$\begin{array}{*{20}{l}}
{f'( – 1)}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{f( – 1 + h) – f( – 1)}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{2{{\left( { – 1 + h} \right)}^2} – 3\left( { – 1 + h} \right) + 5}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{2 – 4h + 2{h^2} + 3 – 3h + 5}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{h\left( {2h – 7} \right)}}{h}} \\
{}& = &{ – 7}
\end{array}$$

$f:x \to {x^3} – 1$, em $a = 0$ e em $a = 1$

$$\begin{array}{*{20}{l}}
{f'(0)}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) – f(0)}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {0 + h} \right)}^3} – 1 – ( – 1)}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{{h^3}}}{h}} \\
{}& = &0
\end{array}$$
$$\begin{array}{*{20}{l}}
{f'(1)}& = &{\mathop {\lim }\limits_{x \to 1} \frac{{f(x) – f(1)}}{{x – 1}}} \\
{}& = &{\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} – 1 – 0}}{{x – 1}}} \\
{}& = &{\mathop {\lim }\limits_{x \to 1} \frac{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{x – 1}}} \\
{}& = &3
\end{array}$$

$f:x \to \frac{1}{{{x^2}}}$, em $a = – 2$

$$\begin{array}{*{20}{l}}
{f'( – 2)}& = &{\mathop {\lim }\limits_{x \to – 2} \frac{{f(x) – f( – 2)}}{{x + 2}}} \\
{}& = &{\mathop {\lim }\limits_{x \to – 2} \frac{{\frac{1}{{{x^2}}} – \frac{1}{4}}}{{x + 2}}} \\
{}& = &{\mathop {\lim }\limits_{x \to – 2} \frac{{\frac{{4 – {x^2}}}{{4{x^2}}}}}{{x – 1}}} \\
{}& = &{\mathop {\lim }\limits_{x \to – 2} \frac{{4\left( {1 + x} \right)\left( {1 – x} \right)}}{{4{x^2}\left( {x – 1} \right)}}} \\
{}& = &{\mathop {\lim }\limits_{x \to – 2} \frac{{ – \left( {1 + x} \right)}}{{{x^2}}}} \\
{}& = &{\frac{1}{4}}
\end{array}$$

$f:x \to \frac{{3x + 2}}{{x – 5}}$, em $a = 4$

$$\begin{array}{*{20}{l}}
{f'(4)}& = &{\mathop {\lim }\limits_{x \to 4} \frac{{f(x) – f(4)}}{{x – 4}}} \\
{}& = &{\mathop {\lim }\limits_{x \to 4} \frac{{\frac{{3x + 2}}{{x – 5}} – ( – 14)}}{{x – 4}}} \\
{}& = &{\mathop {\lim }\limits_{x \to 4} \frac{{\frac{{3x + 2 + 14x – 70}}{{x – 5}}}}{{x – 4}}} \\
{}& = &{\mathop {\lim }\limits_{x \to 4} \frac{{17\left( {x – 4} \right)}}{{\left( {x – 4} \right)\left( {x – 5} \right)}}} \\
{}& = &{ – 17}
\end{array}$$

$f:x \to \sqrt x $, em $a = 4$

$$\begin{array}{*{20}{l}}
{f'(4)}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{f(4 + h) – f(4)}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {4 + h} – 2}}{h}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {4 + h} – 2} \right)\left( {\sqrt {4 + h} + 2} \right)}}{{h\left( {\sqrt {4 + h} + 2} \right)}}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{{4 + h – 4}}{{h\left( {\sqrt {4 + h} + 2} \right)}}} \\
{}& = &{\mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {4 + h} + 2}}} \\
{}& = &{\frac{1}{4}}
\end{array}$$