21 Propiedades de las derivadas

Proposición 21.1.

Sean $\displaystyle f$ y $\displaystyle g$ derivables en $\displaystyle x_{0}$ y $\displaystyle\lambda\in\mathbb{R}$ , entonces:

$\displaystyle f+g$ es derivable en $\displaystyle x_{0}$ y $\displaystyle(f+g)^{\prime}(x_{0})=f^{\prime}(x_{0})+g^{\prime}(x_{0})$ .
$\displaystyle\lambda f$ es derivable en $\displaystyle x_{0}$ y $\displaystyle(\lambda f)^{\prime}(x_{0})=\lambda f^{\prime}(x_{0})$ .
$\displaystyle f\cdot g$ es derivable en $\displaystyle x_{0}$ y $\displaystyle(f\cdot g)^{\prime}(x_{0})=f^{\prime}(x_{0})g(x_{0})+f(x_{0})g^{% \prime}(x_{0})$
Si $\displaystyle g(x_{0})\neq 0,\frac{f}{g}$ es derivable en $\displaystyle x_{0}$ y $\displaystyle(\frac{f}{g})(x_{0})=\frac{f^{\prime}(x_{0})g(x_{0})-f(x_{0})g^{% \prime}(x_{0})}{g^{2}(x_{0})}$ .

Demostración.

	$\displaystyle\displaystyle(f+g)^{\prime}(x_{0})$	$\displaystyle\displaystyle=\mathop{\operator@font l\acute{{\imath}}m}\limits_{% h\to 0}\frac{(f+g)(x_{0}+h)-(f+g)(x_{0})}{h}$
		$\displaystyle\displaystyle=\mathop{\operator@font l\acute{{\imath}}m}\limits_{% h\to 0}\frac{f(x_{0}+h)+g(x_{0}+h)-(f(x_{0})+g(x_{0}))}{h}$
		$\displaystyle\displaystyle=\mathop{\operator@font l\acute{{\imath}}m}\limits_{% h\to 0}\frac{f(x_{0}+h)-f(x_{0})}{h}+\mathop{\operator@font l\acute{{\imath}}m% }\limits_{h\to 0}\frac{g(x_{0}+h)-g(h)}{h}$
		$\displaystyle\displaystyle=f^{\prime}(x_{0})+g^{\prime}(x_{0})$

Como $\displaystyle f,g$ son derivables en $\displaystyle x_{0}\Rightarrow f^{\prime}(x_{0}),g^{\prime}(x_{0})\in\mathbb{R% }\Rightarrow(f+g)^{\prime}(x_{0})=f^{\prime}(x_{0})+g^{\prime}(x_{0})\in% \mathbb{R}$ y $\displaystyle f+g$ es derivable.

	$\displaystyle\displaystyle(\lambda f)^{\prime}(x_{0})$	$\displaystyle\displaystyle=\mathop{\operator@font l\acute{{\imath}}m}\limits_{% h\to 0}\frac{(\lambda f)(x_{0}+h)-(\lambda f)(x_{0})}{h}=\mathop{% \operator@font l\acute{{\imath}}m}\limits_{h\to 0}\frac{\lambda f(x_{0}+h)-% \lambda f(x_{0})}{h}$
		$\displaystyle\displaystyle=\mathop{\operator@font l\acute{{\imath}}m}\limits_{% h\to 0}\lambda\cdot\frac{f(x_{0}+h)-f(x_{0})}{h}=\mathop{\operator@font l% \acute{{\imath}}m}\limits_{h\to 0}\lambda\cdot\mathop{\operator@font l\acute{{% \imath}}m}\limits_{h\to 0}\frac{f(x_{0}+h)-f(x_{0})}{h}$
		$\displaystyle\displaystyle=\lambda f^{\prime}(x_{0})$

$(f\cdot g)^{\prime}(x_{0})=\mathop{\operator@font l\acute{{\imath}}m}\limits_{% h\to 0}\frac{(f\cdot g)(x_{0}+h)-(f\cdot g)(x_{0})}{h}=\mathop{\operator@font l% \acute{{\imath}}m}\limits_{h\to 0}\frac{f(x_{0})g(x_{0}+h)-f(x_{0})g(x_{0})}{h% }=\\ =\mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0}\frac{f(x_{0}+h)g(x% _{0}+h)-f(x_{0})\cdot g(x_{0}+h)+f(x_{0})g(x_{0}+h)-f(x_{0})g(x_{0})}{h}=\\ =\mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0}\frac{g(x_{0}+h)(f(% x_{0}+h)-f(x_{0}))}{h}+\mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0% }\frac{f(x_{0})\cdot(g(x_{0}+h)-g(x_{0}))}{h}=\\ =\mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0}g(x_{0}+h)\cdot% \mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0}\frac{f(x_{0}+h)-f(x% _{0})}{h}+g(x_{0})\cdot\mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0% }\frac{g(x_{0}+h)-g(x_{0})}{h}=\\ =f^{\prime}(x_{0})\cdot\mathop{\operator@font l\acute{{\imath}}m}\limits_{h\to 0% }g(x_{0}+h)+f(x_{0})\cdot g^{\prime}(x_{0})\overset{g\text{ continua en }x_{0}% }{=}f^{\prime}(x_{0})g(x_{0})+f(x_{0})g^{\prime}(x_{0})$
Añadir.

∎

Teorema 21.1.

Sea $\displaystyle f\colon I\to\mathbb{R}$ con $\displaystyle I=(a,b)$ . Si $\displaystyle f$ es derivable en $\displaystyle x_{0}$ , entonces $\displaystyle f$ es continua en $\displaystyle x_{0}$ .

Demostración.

Dado $\displaystyle x\in I$ , $\displaystyle x\neq x_{0}$ ,

\displaystyle f(x)-f(x_{0})=\frac{(f(x)-f(x_{0}))}{x-x_{0}}(x-x_{0})

Además, tenemos que

\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}f(x% )-f(x_{0})=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}\frac% {f(x)-f(x_{0})}{x-x_{0}}\cdot\mathop{\operator@font l\acute{{\imath}}m}\limits% _{x\to x_{0}}x-x_{0}=f^{\prime}(x_{0})\cdot 0=0

Por tanto, $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}(f(% x)-f(x_{0}))=0$ .

\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}f(x% )=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}(f(x)-f(x_{0})% +f(x_{0}))=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}(f(x)% -f(x_{0}))+\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}f(x_{% 0})=f(x_{0})

Luego $\displaystyle f$ es continua en $\displaystyle x_{0}$ por definición. ∎

Teorema 21.2 (Regla de la cadena).

Si $\displaystyle f$ es derivable en $\displaystyle x_{0}$ y $\displaystyle g$ es derivable en $\displaystyle f(x_{0})$ , entonces $\displaystyle g\circ f$ es derivable en $\displaystyle x_{0}$ y

\displaystyle(g\circ f)^{\prime}(x_{0})=g^{\prime}(f(x_{0}))f^{\prime}(x_{0})

Demostración.

No vista en clase. ∎

Ejemplo.

\displaystyle f(x)=\begin{dcases}x^{2}\sin\frac{1}{x}+2x^{2}\quad x\neq 0\\ 0\quad x=0\end{dcases}

1.
$\displaystyle f$ es continua en $\displaystyle x\neq 0$ : $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}f(x% )=f(x_{0})?$ $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}x^{% 2}\sin\frac{1}{x}+2x^{2}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x% \to x_{0}}x^{2}\sin\frac{1}{x}+\mathop{\operator@font l\acute{{\imath}}m}% \limits_{x\to x_{0}}2x^{2}=x^{2}_{0}\sin\frac{1}{x_{0}}+2x^{2}_{0}$ $\displaystyle f$ es continua en $\displaystyle x=0$ ? $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}x^{2}% \sin\frac{1}{x}+2x^{2}\overset{?}{=}\mathop{\operator@font l\acute{{\imath}}m}% \limits_{x\to 0}x^{2}\sin\frac{1}{x}+\underbrace{\mathop{\operator@font l% \acute{{\imath}}m}\limits_{x\to 0}2x^{2}}_{=0}=\mathop{\operator@font l\acute{% {\imath}}m}\limits_{x\to 0}x^{2}\sin\frac{1}{x}=0+0=0$ . $\displaystyle\left|x^{2}\sin\frac{1}{x}\right|=\left|x^{2}\right|\left|\sin% \frac{1}{x}\right|=x^{2}\cdot\left|\sin\frac{1}{x}\right|\leq x^{2}\overset{x% \rightarrow 0}{\longrightarrow}\mathop{\operator@font l\acute{{\imath}}m}% \limits_{x\to 0}\left|x^{2}\sin\frac{1}{x}\right|=0$ . $\displaystyle f$ es derivable en $\displaystyle x=x_{0}\neq 0?$ $\displaystyle f(x)=x^{2}\sin\frac{1}{x}+2x^{2}$ $\displaystyle\frac{1}{x}$ es derivable en $\displaystyle x_{0}\neq 0$ , $\displaystyle\sin(y)$ es derivable en $\displaystyle y=\frac{1}{x_{0}}\overset{R.C.}{\Rightarrow}\sin\frac{1}{x}$ es derivable en $\displaystyle x_{0}$ . Por tanto, $\displaystyle f$ es derivable en $\displaystyle\mathbb{R}\setminus\{0\}$ . $\displaystyle f$ es derivable en $\displaystyle x=0$ ?

$\displaystyle f^{\prime}(x)=\begin{cases}2x\sin\frac{1}{x}+x^{2}\cos\frac{1}{x% }\cdot-\frac{1}{x^{2}}+4x\qquad x\neq 0\\ ??\qquad x=0\end{cases}$

$\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}f^{% \prime}(x)=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}2x\sin% \frac{1}{x}-\cos\frac{1}{x}+4x$ . Veamos que $\displaystyle\cancel{\exists}\mathop{\operator@font l\acute{{\imath}}m}\limits% _{x\to 0}\cos\frac{1}{x}$ . Consideramos $\displaystyle x_{n}=\frac{1}{2\pi n}\Rightarrow\cos\frac{1}{\frac{1}{2\pi n}}=% \cos 2\pi nn=1$ . Tambien $\displaystyle y_{n}=\frac{1}{2\pi n+\pi}\overset{n\rightarrow\infty}{% \longrightarrow}0$ . $\displaystyle\cos\frac{1}{y_{n}}=\cos\frac{1}{\frac{1}{2\pi n+\pi}}=\cos 2\pi n% +\pi=-1$ . Luego

$\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{n\to\infty}2x_% {n}\sin\frac{1}{x_{n}}-\cos\frac{1}{x_{n}}+4x_{n}=-1$

$\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{n\to\infty}2y_% {n}\sin\frac{1}{y_{n}}-\cos\frac{1}{y_{n}}+4y_{n}=1$

Como $\displaystyle 1\neq-1$ , $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}f^{% \prime}(x)$ no existe $\displaystyle\Rightarrow$ $\displaystyle f$ es derivable en $\displaystyle x=0$ ? No sabemos nada. $\displaystyle f^{\prime}(0)=\mathop{\operator@font l\acute{{\imath}}m}\limits_% {h\to 0}\frac{f(0+h)-f(0)}{h}=\mathop{\operator@font l\acute{{\imath}}m}% \limits_{h\to 0}\frac{h^{2}\cdot\sin\frac{1}{h}+2h^{2}}{h}=\mathop{% \operator@font l\acute{{\imath}}m}\limits_{h\to 0}h\sin\frac{1}{h}+2h=0$ (por sandwich). Por tanto, $\displaystyle f$ es derivable en $\displaystyle x=0$ . $\displaystyle f$ es derivable en $\displaystyle\mathbb{R}$ . $\displaystyle f^{\prime}$ es continua en $\displaystyle\mathbb{R}$ :
- $\displaystyle x\neq 0\Rightarrow f^{\prime}$ es continua
- $\displaystyle x=0:$ $\displaystyle\underbrace{\mathop{\operator@font l\acute{{\imath}}m}\limits_{x% \to 0}f^{\prime}(x)}_{\text{No existe}}\overset{?}{=}f^{\prime}(0)=0% \Rightarrow f^{\prime}$ no es continua en $\displaystyle x=0$ .
Su derivada no es continua.

Podemos distinguir varios tipos de funciones según su continuidad y la continuidad de sus derivadas:

Continuas: por ejemplo, $\displaystyle f(x)=\left|x\right|$ . Se dice que $\displaystyle f(x)\in\mathscr{C}^{0},\mathscr{C}$ .
Derivada no continua: por ejemplo, la de Ejemplo: $\displaystyle f(x)$ es continua y derivable pero $\displaystyle f^{\prime}(x)$ no es continua.
Derivada continua: $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}f^{% \prime}(x)=2=f^{\prime}(0)$ . Se dice que $\displaystyle f(x)\in\mathscr{C}^{1}$ . Si la segunda derivada es continua, $\displaystyle f(x)\in\mathscr{C}^{2}$ , etc.

Teorema 21.3 (Teorema del extremo interior).

Sea $\displaystyle c$ un punto interior del intervalo $\displaystyle I$ en el que $\displaystyle f\colon I\to\mathbb{R}$ tiene un extremo relativo. Si la derivada de $\displaystyle f$ en $\displaystyle c$ existe, entonces $\displaystyle f^{\prime}(c)=0$ .

Demostración.

Supongamos que $\displaystyle f$ tiene un máximo relativo en $\displaystyle c\in I$ . La demostración del caso de un mínimo relativo es similar. Si $\displaystyle f^{\prime}(c)>0$ , entonces por el teorema 16.5 existe $\displaystyle\delta>0$ tal que

\displaystyle\frac{f(x)-f(c)}{x-c}>0\text{ para }x\in(c-\delta,c+\delta),x\neq c.

Si $\displaystyle x\in V$ y $\displaystyle x>c$ , se tiene entonces

\displaystyle f(x)-f(c)=(x-c)\cdot\frac{f(x)-f(c)}{x-c}>0

Pero esto contradice la hipótesis de que $\displaystyle f$ tiene un máximo relativo en $\displaystyle c$ . Por tanto, no se puede tener $\displaystyle f^{\prime}(c)>0$ . De manera similar, no se puede tener $\displaystyle f^{\prime}(c)<0$ . Por lo tanto, se debe tener $\displaystyle f^{\prime}(c)=0$ . ∎