Cho hàm số $y = {\log _a}x$. Nếu $0 < a < 1$ thì hàm số:

$\left( {0;1} \right)$

$R$

$R\backslash \left\{ 0 \right\}$

$\left( {0; + \infty } \right)$

$y' = {\log _a}x$

$y' = x\ln a$  

$y' = \dfrac{1}{{x\ln a}}$

$y' = \dfrac{1}{x}\ln a$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{x} = 1$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 - x} \right)}}{x} = 1$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{x} = 1$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln \left( {1 + x} \right)}}{{1 + x}} = 1$

$x = 1$

$y = 0$

$y = 1$          

$x = 0$ 

${y_0} = {\log _a}{x_0}$

${y_0} = x_0^a$       

${y_0} = {a^{{x_0}}}$           

${x_0} = {\log _a}{y_0}$ 

$\left( {1;0} \right)$

$\left( {a,1} \right)$   

$\left( {{a^2};a} \right)$       

$\left( {{a^2};2} \right)$