Derivada
Si $f\left( x \right) = \log x$, entonces:
$f'\left( x \right) = \frac{1}{{\ln 10 \cdot x}}$
o
$\frac{d}{{dx}}\left( {\log x} \right) = \frac{1}{{\ln 10 \cdot x}}$
Demostración
Sea la función $y = f\left( x \right) = \log x$
$y = \log x$
${10^y} = {10^{\log x}}$
${10^y} = x$
${\left[ {{{10}^y}} \right]^\prime } = {\left[ x \right]^\prime }$
$\ln 10 \cdot {10^y} \cdot y' = 1$
$y' = \frac{1}{{\ln 10 \cdot {{10}^y}}}$
$y' = \frac{1}{{\ln 10 \cdot {{10}^{\left( {\log x} \right)}}}}$
$y' = \frac{1}{{\ln 10 \cdot x}} = \frac{1}{{x\ln 10}}$
Aplica el exponencial base 10 en la ecuación
Dado que la exponencial base 10 y logaritmo base 10 son inversas.
Derivación con la regla de la cadena.
Se obtiene la derivada.
Despeja a $y'$.
Sustitute $y$ por $\log x$.
${10^{\left( {\log x} \right)}} = x$ puesto que el exponencial base 10 y el logaritmo base 10 son operaciones inversas.
Cuando el argumento del logaritmo es una función de x distinta de la función identidad, entonces se debe aplicar la regla de la cadena:
Derivada
Sea $f\left( u \right) = \log u$, donde $u$ una función de $x$ derivable; entonces:
$\frac{d}{{dx}}\left( {\log u} \right) = \frac{1}{{\ln 10 \cdot u}}\frac{{du}}{{dx}}$
o
$f'\left( x \right) = \frac{1}{{\ln 10 \cdot u}}u'\left( x \right)$
Deriva las siguientes funciones logarítmicas.
1. $y = \log \left( {{x^4} - 2} \right)$ Sea $u = {x^4} - 2$, entonces $\eqalign{ & \frac{{dy}}{{dx}} = \frac{1}{{\ln 10 \cdot \left( {{x^4} - 2} \right)}}\frac{d}{{dx}}\left( {{x^4} - 2} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln 10 \cdot \left( {{x^4} - 2} \right)}}\left( {4{x^3}} \right) \cr & \frac{{dy}}{{dx}} = \frac{{4{x^3}}}{{\ln 10 \cdot \left( {{x^4} - 2} \right)}} \cr} $ |
2. $y = \log \left( {\log x} \right)$ Sea $u = \log x$, entonces $\frac{{dy}}{{dx}} = \frac{1}{{\ln 10 \cdot \log x}}\frac{d}{{dx}}\left( {\log x} \right)$ $\frac{{dy}}{{dx}} = \frac{1}{{\ln 10 \cdot \log x}}\left( {\frac{1}{{\ln 10 \cdot x}}} \right)$ $\frac{{dy}}{{dx}} = \frac{1}{{\ln {{\left( {10} \right)}^2} \cdot x \cdot \log x}} = \frac{1}{{\ln \left( {10} \right) \cdot x \cdot \ln x}}$ Dado que $\ln \left( x \right) = \ln \left( {10} \right) \cdot \log \left( x \right)$ |
3. $f\left( x \right) = \log \left( {\sin x} \right)$ Sea $u = \sin x$, entonces $\eqalign{ & f'\left( x \right) = \frac{1}{{\ln 10 \cdot \sin x}}\frac{d}{{dx}}\left( {\sin x} \right) \cr & f'\left( x \right) = \frac{1}{{\ln 10 \cdot \sin x}}\left( {\cos x} \right) \cr} $ $\eqalign{ & f'\left( x \right) = \frac{{\cos x}}{{\ln 10 \cdot \sin x}} \cr & f'\left( x \right) = \frac{{\cot x}}{{\ln 10}} \cr} $ |
4. $y = \log \sqrt {1 + 2x} $ $y = \log {\left( {1 + 2x} \right)^{\frac{1}{2}}} = \frac{1}{2}\log \left( {1 + 2x} \right)$ $y' = \frac{1}{2}\frac{1}{{\ln 10 \cdot \left( {1 + 2x} \right)}}\frac{d}{{dx}}\left( {1 + 2x} \right)$ $\eqalign{ & y' = \frac{1}{2}\frac{1}{{\ln 10 \cdot \left( {1 + 2x} \right)}}\left( 2 \right) \cr & y' = \frac{1}{{\ln 10 \cdot \left( {1 + 2x} \right)}} \cr} $ |
Determina la deriva de las siguientes funciones logarítmicas.
1. $y = \log {\left( {x - 2} \right)^2}$
- $\frac{{dy}}{{dx}} = \frac{2}{{\left( {x - 2} \right)\ln \left( {10} \right)}}$
- $\frac{{dy}}{{dx}} = \frac{1}{{\left( {x - 2} \right)\ln \left( {10} \right)}}$
- $\frac{{dy}}{{dx}} = \frac{{\ln \left( {10} \right)}}{{2\left( {x - 2} \right)}}$
$y = 2\log \left( {x - 2} \right)$
Sea $u = x - 2$, entonces
$\eqalign{ & \frac{{dy}}{{dx}} = 2\left[ {\frac{1}{{\left( {x - 2} \right)\ln \left( {10} \right)}}\frac{d}{{dx}}\left( {x - 2} \right)} \right] \cr & \frac{{dy}}{{dx}} = 2\left[ {\frac{1}{{\left( {x - 2} \right)\ln \left( {10} \right)}}\left( 1 \right)} \right] \cr & \frac{{dy}}{{dx}} = \frac{2}{{\left( {x - 2} \right)\ln \left( {10} \right)}} \cr} $
2. $y = \log \frac{1}{x}$
- $y' = \frac{{\ln 10}}{x}$
- $y' = \frac{x}{{\ln 10}}$
- $y' = \frac{1}{{\ln 10 \cdot x}}$
$y = \log 1 - \log x$
$y' = \frac{d}{{dx}}\left( {1 - \log x} \right) = \frac{d}{{dx}}\left( 1 \right) - \frac{d}{{dx}}\left( {\log x} \right)$
$y' = \frac{1}{{\ln 10 \cdot x}}$
3. $y = \log \left( {\sec x} \right)$
- $y' = \frac{{\tan x}}{{\ln 10 \cdot \sec x}}$
- $y' = \frac{{\tan x}}{{\ln 10}}$
- $y' = \frac{{{{\tan }^2}x}}{{\ln 10}}$
$y' = \frac{1}{{\ln 10 \cdot \sec x}} \cdot \frac{d}{{dx}}\left( {\sec x} \right)$
$y' = \frac{1}{{\ln 10 \cdot \sec x}} \cdot \left( {\tan x\sec x} \right)$
$y' = \frac{{\tan x}}{{\ln 10}}$