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<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
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<title>Derivadas</title>
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<h1 class="title">Derivadas</h1>
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<section>
<header>
<h1 id="concepto-de-derivabilidad">Concepto de derivabilidad</h1>
</header>
<p>Sea <span class="math">\(I\)</span> un intervalo de <span class="math">\(\mathbb{R}\)</span>. Una función <span class="math">\(f \colon I \subset \mathbb{R} \rightarrow \mathbb{R}\)</span> es <em>derivable</em> en <span class="math">\(a \in I\)</span> si existe <span class="math">\( \lim_{x \to a}\, \frac{f(x)-f(a)}{x-a}\)</span>.</p>
<p>A dicho límite lo notaremos <span class="math">\(f'(a)\)</span>. A la función <span class="math">\(a \mapsto f'(a)\)</span> la llamaremos <em>función derivada</em> de <span class="math">\(f\)</span> y la notaremos <span class="math">\(f'\)</span>.</p>
<p>La condición de ser derivable es más fuerte que la de ser continua, ya que si <span class="math">\(f \colon I \to \mathbb{R}\)</span> es derivable en <span class="math">\(a\in I\)</span>, entonces <span class="math">\(f\)</span> es continua en <span class="math">\(a\)</span>. El recíproco no es cierto. Hay funciones continuas que no son derivables.</p>
<p><span class="math">\( \)</span></p>
<ol>
<li><p>Si <span class="math">\(f\)</span> es una función constante (<span class="math">\(f(x)=c, \ \forall x \in I\)</span>), entonces su derivada es siempre cero. En efecto: <span class="math">\[f'(a)= \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{x \to a}
\frac{c-c}{x-a} = \lim_{x \to a} \frac{0}{x-a}= 0.\]</span> Concluimos entonces que, si <span class="math">\(f(x)=c , \ \forall x \in I \)</span>, entonces <span class="math">\(f'(x)=0\)</span>, <span class="math">\(\forall x \in I\)</span>.</p></li>
<li><p>La derivada de la función identidad (<span class="math">\(f(x)=x\)</span>) es igual a <span class="math">\(1\)</span> en cualquier punto. Lo comprobamos aplicando la definición de derivada: <span class="math">\[f'(a)= \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{x \to a}
\frac{x-a}{x-a} = 1.\]</span> Por tanto, si <span class="math">\(f(x)=x\)</span>, entonces <span class="math">\(f'(x)=1\)</span>.</p></li>
<li><p>La función <span class="math">\(f(x)=x^2\)</span> es derivable. Su derivada en un punto <span class="math">\(a\)</span> es, según la definición, <span class="math">\[f'(a)= \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{x \to a}
\frac{x^{2}-a^{2}}{x-a} = \lim_{x \to a} \frac{(x+a)(x-a)}{x-a}= 2a.\]</span> Obtenemos así la fórmula usual de la derivada de <span class="math">\(f(x)=x^2\)</span>, esto es, que <span class="math">\(f'(x)=2x\)</span>.</p></li>
</ol>
<h2 id="derivadas-laterales">Derivadas laterales</h2>
<p>Puesto que la derivada está definida como un límite y sabemos la relación entre límites laterales y límite, podemos hablar de <em>derivadas laterales.</em></p>
<p>Sea <span class="math">\(f \colon I \to \mathbb{R}\)</span>, <span class="math">\(a \in I\)</span>, de forma que <span class="math">\(\{ x \in I : x <a \} \ne \emptyset\)</span>. Se dice que <span class="math">\(f\)</span> es <em>derivable por la izquierda</em> en el punto <span class="math">\(a\)</span> si existe <span class="math">\( \lim_{x \to a^{-}} \frac{f(x)-f(a)}{x-a} = f'(a^{-})\)</span>. Este límite se llama <em>derivada lateral izquierda</em> de <span class="math">\(f\)</span> en el punto <span class="math">\(a\)</span>.</p>
<p>Si ahora el punto <span class="math">\(a\)</span> es tal que <span class="math">\(\{ x \in I : x > a \} \ne \emptyset\)</span>, se dice que <span class="math">\(f\)</span> es <em>derivable por la derecha</em> en el punto <span class="math">\(a\)</span> si existe <span class="math">\(\lim_{x \to a^+} \frac{f(x)-f(a)}{x-a} = f'(a^+)\)</span>. Este límite se llama <em>derivada lateral derecha</em> de <span class="math">\(f\)</span> en el punto <span class="math">\(a\)</span>.</p>
<p>Para que una función sea derivable deben de existir todas las derivadas laterales que tengan sentido y coincidir.</p>
<p>Consideremos la función <span class="math">\(f(x)=\lvert x\rvert\)</span>, <span class="math">\(\forall \, x \in \mathbb{R}\)</span>. Vamos a estudiar la derivabilidad de la función valor absoluto en el punto <span class="math">\(a=0\)</span>. Para ello, calculamos las derivadas laterales en dicho punto:</p>
<p><span class="math">\[\begin{aligned}
f'(0^+) &= \lim_{x \to 0^+} \frac{f(x)-f(0)}{x-0}= \lim_{x \to 0^+} \frac{|x|}{x}= \lim_{x \to 0^+} \frac{x}{x}= 1, \\
f'(0^-) &= \lim_{x \to 0^-} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0^-} \frac{|x|}{x}= \lim_{x \to 0^-} \frac{-x}{x}= -1.\end{aligned}\]</span></p>
<p>Por tanto, la función valor absoluto no es derivable en cero, ya que las derivadas laterales en dicho punto no coinciden; sin embargo, recordemos que esta función sí es continua en cero.</p>
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</section>
<section>
<header>
<h1 id="reglas-de-derivabilidad">Reglas de derivabilidad</h1>
</header>
<p>Sean <span class="math">\(f,g \colon I \rightarrow \mathbb{R}\)</span> funciones derivables en <span class="math">\(I\)</span>. Entonces</p>
<ul>
<li><p>Derivada de la suma: <span class="math">\(
(f+g)'(x)= f'(x)+g'(x).
\)</span></p></li>
<li><p>Derivada del producto: <span class="math">\(
(fg)'(x)= f'(x)g(x)+f(x)g'(x).
\)</span></p></li>
<li><p>Derivada del cociente: <span class="math">\(
\left(\dfrac{f}{g}\right)'(x)= \displaystyle \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^{2}}.
\)</span></p></li>
<li><p>(Regla de la cadena) Derivada de la composición: <span class="math">\((g\circ f)'(x)=g'(f(x) \, f'(x)\)</span></p></li>
<li><p>Si <span class="math">\(f(x)=x^n\)</span>, entonces, <span class="math">\(f'(x)=nx^{n-1}\)</span>.</p></li>
</ul>
<p>Vamos a calcular la derivada de cada una de las siguientes funciones:</p>
<ol>
<li><p><span class="math">Sea \(f(x)=7x^3-2x^2+4x-15\). Entonces \(f'(x)= 21x^2-4x+4\)</span>.</p></li>
<li><p><span class="math">Para \(f(x)=\frac{x^2-1}{x}\), tenemos \(f'(x) = \frac{2x\cdot x-(x^2-1)}{x^2}
= \frac{x^2+1}{x^2}\)</span>.</p></li>
<li><p>Se considera la función a trozos: <span class="math">\[f(x)= \begin{cases}
x-4, & \text{si } x <0, \\
2x^2-1, & \text{si } 0 \le x \le 1, \\
\dfrac{1}{x}\, , & \text{si } x>1.
\end{cases}\]</span> En los intervalos <span class="math">\(]-\infty,0[ \cup ]0,1[ \cup ]1,+\infty[\)</span> el cálculo de la derivada se realiza aplicando las reglas anteriores. </p>
<p><span class="math">\[\begin{aligned}
& \text{Si } x <0, \text{entonces } \ f'(x)=1. \\
& \text{Si } 0 < x <1, \text{entonces } \ f'(x)=4x. \\
& \text{Si } x>1, \text{entonces } \ f'(x)=\frac{-1}{x^2}.
\end{aligned}\]</span></p>
<p>Ahora hay que analizar la derivabilidad en los puntos <span class="math">\(x=0\)</span> y <span class="math">\(x=1\)</span>. En primer lugar descartamos el punto <span class="math">\(x=0\)</span> como punto de derivabilidad ya que la función no es continua en dicho punto (basta con comprobar que los límites laterales en el punto <span class="math">\(0\)</span> son distintos). En el punto <span class="math">\(1\)</span> sí hay continuidad, así que estudiamos las derivadas laterales para comprobar que existen y coinciden.</p>
<p><span class="math">\[\begin{aligned}
f'(1^-) &=\lim_{x \to 1^-} \frac{f(x)-f(1)}{x-1}=\lim_{x \to 1^-}\frac{2x^2-2}{x-1}=
\lim_{x \to 1^-} \frac{2(x-1)(x+1)}{x-1}=\lim_{x \to 1^-} 2(x+1) = 4.\\
f'(1^+) &=\lim_{x \to 1^+} \frac{f(x)-f(1)}{x-1}=\lim_{x \to 1^+}\frac{1/x-1}{x-1}=
\lim_{x \to 1^+} \frac{1-x}{x(x-1)}=\lim_{x \to 1^+}\frac{-1}{x}=-1.
\end{aligned}\]</span></p>
<p>Las dos derivadas laterales en <span class="math">\(1\)</span> existen, pero no coinciden, así que la función <span class="math">\(f\)</span> es derivable en <span class="math">\(\mathbb{R} \setminus \{0,1\}\)</span>.</p></li>
<li><p>Sea <span class="math">\(f(x)=(x^3-2x+4)^{17}\)</span>. Para calcular su derivada, a nadie se le ocurre desarrollar esta potencia, ¿verdad? Vamos a hacer uso de la regla de la cadena ya que la función no es más que la composición del polinomio <span class="math">\(x^3-2x+4\)</span> con la potencial <span class="math">\(x^{17}\)</span>. Así que aplicando dicha regla nos queda: <span class="math">\(f'(x)=17(x^3-2x+4)^{16}(3x^2-2)\)</span>.</p></li>
</ol>
<article>
<header>
<h3>Ejercicio</h3>
<p>Calcula la derivada de las siguientes funciones</p>
</header>
<ol>
<li><p><span class="math">\( f(x)=4x^3-5x^2-7x+1\)</span>,
<button id="e1-1" class="button" onclick="show2('e1-1');">Solución</button>
<div id="sol-e1-1" style="display:none;">
\(f'(x)=12x^2-10x-7\),</div></p></li>
<li><p><span class="math">\(f(x)=x^2+\dfrac{1}{x}\)</span>,
<button id="e1-2" class="button" onclick="show2('e1-2');">Solución</button>
<div id="sol-e1-2" style="display:none;">
\(f'(x)=2x-\dfrac{1}{x^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=5\sqrt{x}\)</span>,
<button id="e1-3" class="button" onclick="show2('e1-3');">Solución</button>
<div id="sol-e1-3" style="display:none;">
\(f'(x)=\dfrac{5}{2\sqrt{x}}\),</div></p></li>
<li><p><span class="math">\(f(x)=(4x^2-3x+8)^{10}\)</span>,
<button id="e1-4" class="button" onclick="show2('e1-4');">Solución</button>
<div id="sol-e1-4" style="display:none;">
\(f'(x)=10 (8 x - 3) (4 x^2 - 3 x + 8)^9\),</div></p></li>
<li><p><span class="math">\(f(x)=(x^3-4x+5)^{1/2}\)</span>,
<button id="e1-5" class="button" onclick="show2('e1-5');">Solución</button>
<div id="sol-e1-5" style="display:none;">
\(f'(x)=\dfrac{3x^2-4}{2\sqrt{x^3-4x+5}}\),</div></p></li>
<li><p><span class="math">\(f(x)=\sqrt[4]{x^3}\)</span>,
<button id="e1-6" class="button" onclick="show2('e1-6');">Solución</button>
<div id="sol-e1-6" style="display:none;">
\(f'(x)=\dfrac{3x^2}{4\sqrt[4]{x^9}}\),</div></p></li>
<li><p><span class="math">\( f(x)=\dfrac{1}{x^2-3x+4}\)</span>,
<button id="e1-7" class="button" onclick="show2('e1-7');">Solución</button>
<div id="sol-e1-7" style="display:none;">
\(f'(x)=\dfrac{3 - 2 x}{(x^2 - 3 x + 4)^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\dfrac{x^2}{x+1}\)</span>,
<button id="e1-8" class="button" onclick="show2('e1-8');">Solución</button>
<div id="sol-e1-8" style="display:none;">
\(f'(x)=\dfrac{x (x + 2)}{(x + 1)^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\dfrac{x^2+1}{x-1}\)</span>,
<button id="e1-9" class="button" onclick="show2('e1-9');">Solución</button>
<div id="sol-e1-9" style="display:none;">
\(f'(x)=\dfrac{x^2 - 2 x - 1}{(x - 1)^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\dfrac{x+5}{(x^2-1)^2}\)</span>,
<button id="e1-10" class="button" onclick="show2('e1-10');">Solución</button>
<div id="sol-e1-10" style="display:none;">
\(f'(x)=\dfrac{3x^2+20x+1}{(x^2-1)^3}\),</div></p></li>
<li><p><span class="math">\(f(x)=x^2+\dfrac{1}{\sqrt{x}}\)</span>,
<button id="e1-11" class="button" onclick="show2('e1-11');">Solución</button>
<div id="sol-e1-11" style="display:none;">
\(f'(x)=2x-\dfrac{1}{2\sqrt{x^3}}\),</div></p></li>
<li><p><span class="math">\(f(x)=\dfrac{\sqrt{x+1}}{x}\).</span>
<button id="e1-12" class="button" onclick="show2('e1-12');">Solución</button>
<div id="sol-e1-12" style="display:none;">
\(f'(x)=\dfrac{-x - 2}{2 x^2 \sqrt{x + 1}}\).</div></p></li>
</ol>
</article>
</section>
<section>
<header>
<h1 id="derivación-de-funciones-elementales">Derivación de funciones elementales</h1>
</header>
<p>Usando las reglas que hemos visto y conociendo la derivada de las funciones elementales, podemos calcular la derivada de cualquier función. En la siguiente tabla tenemos la derivada (y primitiva) de algunas funciones usuales.</p>
<style>
tr.border_bottom td {
border-bottom:1pt solid black;
}
</style>
<table>
<tr class="border_bottom"><td><span class="math">\(f(x)\)</span> </td><td> <span class="math">\(f'(x)\)</span> </td><td> <span class="math">\(\int f(x)\,\mathrm{d}x\)</span></tr>
<tr><td><span class="math">\(x^n\)</span> </td><td> <span class="math">\(nx^{n-1}\)</span> </td><td> <span class="math">\((n+1)^{-1}x^{n+1}\)</span></tr>
<tr><td><span class="math">\(1/x\)</span> </td><td> <span class="math">\(-x^{-2}\)</span> </td><td> <span class="math">\(\ln (\lvert x\rvert )\)</span></tr>
<tr><td><span class="math">\(a^x\)</span> </td><td> <span class="math">\(a^x\ln(a)\)</span> </td><td> <span class="math">\(a^x/\ln(a)\)</span></tr>
<tr><td><span class="math">\(e^x\)</span> </td><td> <span class="math">\(e^x\)</span> </td><td> <span class="math">\(e^x\)</span></tr>
<tr><td><span class="math">\(\ln(x)\)</span> </td><td> <span class="math">\(1/x\)</span> </td><td> <span class="math">\(x\ln(x)-x\)</span></tr>
<tr><td><span class="math">\(\operatorname{sen}(x)\)</span> </td><td> <span class="math">\(\cos(x)\)</span> </td><td> <span class="math">\(-\cos(x)\)</span></tr>
<tr><td><span class="math">\(\cos(x)\)</span> </td><td> <span class="math">\(-\operatorname{sen}(x)\)</span> </td><td> <span class="math">\(\operatorname{sen}(x)\)</span></tr>
<tr><td><span class="math">\(\tan(x)\)</span> </td><td> <span class="math">\(\cos^{-2}(x)\)</span> </td><td> <span class="math">\(-\ln (\lvert\cos(x)\rvert )\)</span></tr>
<tr><td><span class="math">\(\operatorname{senh}(x)\)</span> </td><td> <span class="math">\(\cosh(x)\)</span> </td><td> <span class="math">\(\cosh(x)\)</span></tr>
<tr><td><span class="math">\(\cosh(x)\)</span> </td><td> <span class="math">\(\operatorname{senh}(x)\)</span> </td><td> <span class="math">\(\operatorname{senh}(x)\)</span></tr>
<tr><td><span class="math">\(\sec(x)\)</span> </td><td> <span class="math">\(\operatorname{sen}(x)/\cos^2(x)\)</span> </td><td></tr>
<tr><td><span class="math">\(\operatorname{cosec}(x)\)</span> </td><td> <span class="math">\(-\cos(x)/\operatorname{sen}^2(x)\)</span> </td><td></tr>
<tr><td><span class="math">\(\operatorname{arcsen}(x)\)</span> </td><td><span class="math">\(\frac{1}{\sqrt{1-x^2}}\)</span> </td><td></tr>
<tr><td><span class="math">\(\arccos(x)\)</span> </td><td><span class="math">\(\frac{-1}{\sqrt{1-x^2}}\)</span></td><td></tr>
<tr><td><span class="math">\(\arctan(x)\)</span> </td><td><span class="math">\(\frac{1}{1+x^2}\)</span> </td><td></tr>
</table>
<article>
<header>
<h3>Ejercicio</h3>
<p>Calcula la derivada de las siguientes funciones</p>
</header>
<ol>
<li><p><span class="math">\( f(x)=\ln(5x)\)</span>,
<button id="e2-1" class="button" onclick="show2('e2-1');">Solución</button>
<div id="sol-e2-1" style="display:none;"> \(f'(x)=-\dfrac{1}{x}\),</div></p></li>
<li><p><span class="math">\(f(x)=e^{1/x}\)</span>,
<button id="e2-2" class="button" onclick="show2('e2-2');">Solución</button>
<div id="sol-e2-2" style="display:none;"> \(f'(x)=-\dfrac{e^{1/x}}{x^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\ln \left(\sqrt{\cos(x)} \right)\)</span>,
<button id="e2-3" class="button" onclick="show2('e2-3');">Solución</button>
<div id="sol-e2-3" style="display:none;"> \(f'(x)=-\dfrac{\tan(x)}{2}\),</div></p></li>
<li><p><span class="math">\(f(x)=e^{\sqrt{x^2+1}}\)</span>,
<button id="e2-4" class="button" onclick="show2('e2-4');">Solución</button>
<div id="sol-e2-4" style="display:none;"> \(f'(x)=\dfrac{e^{\sqrt{x^2+1}}x}{\sqrt{x^2+1}}\),</div></p></li>
<li><p><span class="math">\(f(x)=\ln(\sqrt{x})\)</span>,
<button id="e2-5" class="button" onclick="show2('e2-5');">Solución</button>
<div id="sol-e2-5" style="display:none;"> \(f'(x)=\dfrac{1}{2x}\),</div></p></li>
<li><p><span class="math">\(f(x)=2^{x^3-2x+1}\)</span>,
<button id="e2-6" class="button" onclick="show2('e2-6');">Solución</button>
<div id="sol-e2-6" style="display:none;">
\(f'(x)=2^{x^3 - 2 x + 1} (3 x^2 - 2) \ln(2)\),</div></p></li>
<li><p><span class="math">\( f(x)=\operatorname{sen}(4x^2-1)\)</span>,
<button id="e2-7" class="button" onclick="show2('e2-7');">Solución</button>
<div id="sol-e2-7" style="display:none;">
\(f'(x)=8 x \cos(1 - 4 x^2)\),</div></p></li>
<li><p><span class="math">\(f(x)=\operatorname{sen} (\cos(x))\)</span>,
<button id="e2-8" class="button" onclick="show2('e2-8');">Solución</button>
<div id="sol-e2-8" style="display:none;"> \(f'(x)=\operatorname{sen}(x) (-\cos(\cos(x)))\),</div></p></li>
<li><p><span class="math">\(f(x)=\cos(\operatorname{sen} (x))\)</span>,
<button id="e2-9" class="button" onclick="show2('e2-9');">Solución</button>
<div id="sol-e2-9" style="display:none;">
\(f'(x)= \operatorname{sen}(\operatorname{sen}(x)) (-\cos(x))\),</div></p></li>
<li><p><span class="math">\(f(x)=\tan \left(\frac{1}{x} \right)\)</span>,
<button id="e2-10" class="button" onclick="show2('e2-10');">Solución</button>
<div id="sol-e2-10" style="display:none;">
\(f'(x)= -\dfrac{\sec^2(1/x)}{x^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\cos(e^x)\)</span>,
<button id="e2-11" class="button" onclick="show2('e2-11');">Solución</button>
<div id="sol-e2-11" style="display:none;">
\(f'(x)= -e^x \operatorname{sen}(e^x)\),</div></p></li>
<li><p><span class="math">\(f(x)=\tan(\ln(x))\)</span>,
<button id="e2-12" class="button" onclick="show2('e2-12');">Solución</button>
<div id="sol-e2-12" style="display:none;">
\(f'(x)= \dfrac{\sec^2(\ln(x))}x\),</div></p></li>
<li><p><span class="math">\( f(x)=\operatorname{arcsen} (3x)\)</span>,
<button id="e2-13" class="button" onclick="show2('e2-13');">Solución</button>
<div id="sol-e2-13" style="display:none;">
\(f'(x)= \dfrac{3}{\sqrt{1 - 9 x^2}}\),</div></p></li>
<li><p><span class="math">\(f(x)=\arccos(\sqrt{x})\)</span>,
<button id="e2-14" class="button" onclick="show2('e2-14');">Solución</button>
<div id="sol-e2-14" style="display:none;">
\(f'(x)= -\dfrac{1}{2 \sqrt{-(x - 1) x}}\),</div></p></li>
<li><p><span class="math">\(f(x)=\arctan(x^2-1)\)</span>,
<button id="e2-15" class="button" onclick="show2('e2-15');">Solución</button>
<div id="sol-e2-15" style="display:none;">
\(f'(x)= \dfrac{2 x}{x^4 - 2 x^2 + 2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\operatorname{arcsen} \left(\frac{1}{x} \right)\)</span>,
<button id="e2-16" class="button" onclick="show2('e2-16');">Solución</button>
<div id="sol-e2-16" style="display:none;">
\(f'(x)= -\dfrac{1}{\sqrt{1 - 1/x^2} x^2}\),</div></p></li>
<li><p><span class="math">\(f(x)=\arccos (e^x)\)</span>,
<button id="e2-17" class="button" onclick="show2('e2-17');">Solución</button>
<div id="sol-e2-17" style="display:none;">
\(f'(x)= -\dfrac{e^x}{\sqrt{1 - e^{2 x}}}\),</div></p></li>
<li><p><span class="math">\( f(x)=\sqrt{\ln(3+\cos(x^2+1))}\)</span>,
<button id="e2-18" class="button" onclick="show2('e2-18');">Solución</button>
<div id="sol-e2-18" style="display:none;">
\(f'(x)=-\dfrac{x \mathrm{sen}(x^2 + 1)}{(\cos(x^2 + 1) + 3) \sqrt{\ln(\cos(x^2 + 1) + 3)}}\),</div></p></li>
<li><p><span class="math">\(f(x)=\arctan \left(\ln(x) \right)\).</span>
<button id="e2-19" class="button" onclick="show2('e2-19');">Solución</button>
<div id="sol-e2-19" style="display:none;">
\(f'(x)= \dfrac{ 1}{x + x \ln^2(x)}\).</div></p></li>
</ol>
</article>
<p>Si repasamos lo hecho, nos damos cuenta de que nos ha quedado sin resolver el cálculo de la derivada cuando aparecen funciones tanto en la base como en el exponente. Aunque existe una regla para su cálculo, preferimos hacer el desarrollo completo, lo que se suele conocer como derivación logarítmica.</p>
<h2 id="derivación-logarítmica">Derivación logarítmica</h2>
<p>La clave para calcular la derivada de <span class="math">\(f(x)^{g(x)}\)</span> está en derivar no dicha función sino su logaritmo: por abreviar, si escribimos <span class="math">\(y(x)=f(x)^{g(x)}\)</span>, entonces <span class="math">\[\ln(y(x)) = \ln\left( f(x)^{g(x)} \right) = g(x) \ln (f(x)),\]</span> derivamos usando la regla para derivar un producto, <span class="math">\[\left( \ln(y) \right)^{\prime} = \frac{y'(x)}{y(x)} = g'(x)\ln(f(x)) + g(x)\, \frac{f'(x)}{f(x)} \, ,\]</span> y despejamos <span class="math">\(y'(x)\)</span></p>
<p><span class="math">\[\begin{aligned}
y'(x) & = y(x) \left( g'(x)\ln(f(x)) + g(x)\, \frac{f'(x)}{f(x)}\right) \\
& = f(x)^{g(x)} \left( g'(x)\ln(f(x)) + g(x)\, \frac{f'(x)}{f(x)} \right) .\end{aligned}\]</span></p>
<p>Vamos a calcular la derivada de la función <span class="math">\(y(x)=(x^2+x)^{\operatorname{sen}(x)}\)</span>.</p>
<p>Tomamos logaritmos <span class="math">\[\ln (y(x)) = \ln \left( (x^2+x)^{\operatorname{sen}(x)} \right) = \operatorname{sen}(x) \ln (x^2+x),\]</span></p>
<p>derivamos <span class="math">\[\ln(y(x))' = \frac{y'(x)}{y(x)} = \cos(x) \ln(x^2+x) + \operatorname{sen}(x)\frac{2x+1}{x^2+x}\]</span></p>
<p>y despejamos <span class="math">\(y'(x)\)</span></p>
<p><span class="math">\[\begin{aligned}
y'(x) & = y(x) \left( \cos(x) \ln(x^2+x) + \operatorname{sen}(x)\frac{2x+1}{x^2+x} \right) \\
& = (x^2+x)^{\operatorname{sen}(x)}\left( \cos(x) \ln(x^2+x) + \operatorname{sen}(x)\frac{2x+1}{x^2+x} \right) .\end{aligned}\]</span></p>
<article>
<header>
<h3>Ejercicio</h3>
<p>Calcula la derivada de las siguientes funciones</p>
</header>
<ol>
<!--<li><p><span class="math">\(f(x)=\operatorname{arcsen} \left(\sqrt{1-x^2}\right)\)</span>,</p></li>
<li><p><span class="math">\(f(x)=\left(4-\cos(2x)\right)\, \tan(x)\)</span>,</p></li>
<li><p><span class="math">\(f(x)=\ln \left(1+\sqrt{x^2+1}\right)\)</span>,</p></li>-->
<li><p><span class="math">\(f(x)=x^{x}\)</span>,
<button id="e3-1" class="button" onclick="show2('e3-1');">Solución</button>
<div id="sol-e3-1" style="display:none;">
\(f'(x)= x^x (\ln(x) + 1)\),</div></p></li>
<li><p><span class="math">\(f(x)=(x^2+1)^{x^2}\).</span>
<button id="e3-2" class="button" onclick="show2('e3-2');">Solución</button>
<div id="sol-e3-2" style="display:none;">
\(f'(x)= 2 x (x^2 + 1)^{x^2 - 1} (x^2 + (x^2 + 1) \ln(x^2 + 1))\).</div></p></li>
</ol>
</article>
</section>
<hr>
<p style="font-size: 10pt">Esta página está basada en las transparencias de Jerónimo Alaminos Prats, José Extremera Lizana y Pilar Muñoz Rivas para el Curso Cero de la ETSIIT de la Universidad de Granada. Las representaciones gráficas se han realizado con <a href="http://jsxgraph.uni-bayreuth.de">JSXGraph</a>. El maquetado de la página ha sido realizado por Pedro A. García Sánchez.</p>
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