\n\\[\\begin{array}{*{20}{c}}
\n{f\\left( x \\right)}& = &{\\left\\{ {\\begin{array}{*{20}{c}}
\nx& \\Leftarrow &{x > 0} \\\\
\n{ – {x^2}}& \\Leftarrow &{x \\leqslant 0}
\n\\end{array}} \\right.}
\n\\end{array}\\]<\/p>\n<\/blockquote>\n
Calculemos as derivadas laterais no ponto $x = 0$:<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{\\begin{array}{*{20}{l}}
\n{f’\\left( {{0^ – }} \\right)}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ – }} \\frac{{f\\left( {0 + h} \\right) – f\\left( 0 \\right)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ – }} \\frac{{ – {{\\left( {0 + h} \\right)}^2} – 0}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ – }} \\frac{{ – {h^2}}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ – }} \\left( { – h} \\right)} \\\\
\n{}& = &0
\n\\end{array}}&{}&{}&{}&{\\begin{array}{*{20}{l}}
\n{f’\\left( {{0^ + }} \\right)}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ + }} \\frac{{f\\left( {0 + h} \\right) – f\\left( 0 \\right)}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ + }} \\frac{{\\left( {0 + h} \\right) – 0}}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ + }} \\frac{h}{h}} \\\\
\n{}& = &{\\mathop {\\lim }\\limits_{h \\to {0^ + }} \\left( 1 \\right)} \\\\
\n{}& = &1
\n\\end{array}}
\n\\end{array}\\]<\/p>\n
Como $f’\\left( {{0^ – }} \\right) \\ne f’\\left( {{0^ + }} \\right)$, ent\u00e3o a fun\u00e7\u00e3o $f$ n\u00e3o tem derivada no ponto de abcissa $0$, apesar de cont\u00ednua.<\/p>\n<\/p>\n