Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
y = f\left( x \right) = \frac{2}{x}\\
f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{2}{x} - \frac{2}{{{x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{2\left( {{x_0} - x} \right)}}{{x.{x_0}}}}}{{x - {x_0}}}\\
= \mathop {\lim }\limits_{x \to {x_0}} \frac{{2\left( {{x_0} - x} \right)}}{{x.{x_0}.\left( {x - {x_0}} \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - 2}}{{x.{x_0}}} = \frac{{ - 2}}{{{x_0}.{x_0}}} = \frac{{ - 2}}{{{x_0}^2}}\\
b,\\
y = \sqrt {x + 5} \\
f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sqrt {x + 5} - \sqrt {{x_0} + 5} }}{{x - {x_0}}}\\
= \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {\sqrt {x + 5} - \sqrt {{x_0} + 5} } \right)\left( {\sqrt {x + 5} + \sqrt {{x_0} + 5} } \right)}}{{\left( {x - {x_0}} \right)\left( {\sqrt {x + 5} + \sqrt {{x_0} + 5} } \right)}}\\
= \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x + 5} \right) - \left( {{x_0} + 5} \right)}}{{\left( {x - {x_0}} \right)\left( {\sqrt {x + 5} + \sqrt {{x_0} + 5} } \right)}}\\
= \mathop {\lim }\limits_{x \to {x_0}} \frac{1}{{\sqrt {x + 5} + \sqrt {{x_0} + 5} }}\\
= \frac{1}{{\sqrt {{x_0} + 5} + \sqrt {{x_0} + 5} }} = \frac{1}{{2\sqrt {{x_0} + 5} }}
\end{array}\)
