El truco consiste, ni más ni menos, que en obtener la función derivada de f usando las reglas de derivación, en cada trozo, pero excluyendo los "bordes" del dominio de cada trozo, es decir, los puntos de cambio de trozo. Después bastará con comprobar que ambas derivadas "pegan bien", es decir, la recién obtenida función es continua.
Llegados aquí, pueden pasar dos cosas.
1. Si resulta que la función f' puede extenderse a continua (porque los trozos derivados pegan bien), pues afirmamos que la función f es derivable.
2. Si los trozos derivados no pegan bien, decimos que f no es derivable en ese punto.
Pues bien, lo afirmado en el punto 2 no es correcto. Imaginemos la función:
Es decir, hay funciones derivables cuya derivada no es una función continua (es decir, los trozos de su derivada no pegan bien pero sí es derivable). Se dice que son funciones derivables pero no de clase $C^1$. Más aquí.
Entonces, ¿tenemos que renunciar a este maravilloso truco? ¿Hay que usar siempre la definición de derivada? No todo está perdido. Si los trozos no pegan bien pero los límites existen y son finitos, entonces sí podemos afirmar que la función no es derivable.
Veámoslo:
Sean $g$ y $h$ funciones continuas y derivables, y sea:
$
f ( x ) = \left\{ \begin{array} { l l } { g( x) ) } & { x < c } \\ { h ( x ) } & { x \geq c } \end{array} \right.
$
Entonces, la derivada será:
$
f '( x ) = \left\{ \begin{array} { l l } { g'( x) ) } & { x < c } \\ { h' ( x ) } & { x > c } \end{array} \right.
$
No sabemos si $f$ es, a priori, derivable en $c$ o no. Lo será si $\lim _ { x \rightarrow c } \frac { f ( x ) - f \left(c \right) } { x -c }$ existe y es finito, es decir, existen:
$
L_1=\lim _ { x \rightarrow c^{-} } \frac { f ( x ) - f \left(c \right) } { x -c }=\lim _ { x \rightarrow c^{-} } \frac { g( x ) - h \left(c \right) } { x -c }
$
y
$
L_2=\lim _ { x \rightarrow c^{+} } \frac { f ( x ) - f \left(c \right) } { x -c }=\lim _ { x \rightarrow c^{-} } \frac { h( x ) - h \left(c \right) } { x -c }
$
y además son iguales y finitos. En caso contrario, no es derivable.
Supongamos que $\lim _ { x \rightarrow c^{-} }g'(x)=a \neq b=lim _ { x \rightarrow c^{+} }h'(x)$. Entonces, aplicando la regla de L'Hopital:
$
L_1=\lim _ { x \rightarrow c^{-} } \frac { g( x ) - h \left(c \right) } { x -c }=\lim _ { x \rightarrow c^{-} } \frac { g'( x ) } { 1}=a
$
y
$
L_2=\lim _ { x \rightarrow c^{+} } \frac { h( x ) - h \left(c \right) } { x -c }=\lim _ { x \rightarrow c^{+} } \frac { h'( x ) } { 1}=b
$
Por tanto, $L_1 \neq L_2$ y, por tanto, $f$ no es derivable en $c$.
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