Giải thích các bước giải:

Câu 1:

\(\begin{array}{l}
f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{3 - \sqrt {4 - x} }}{4} - \frac{1}{4}}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{2 - \sqrt {4 - x} }}{{4x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - \sqrt {4 - x} } \right)\left( {2 + \sqrt {4 - x} } \right)}}{{4x\left( {2 + \sqrt {4 - x} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{{2^2} - \left( {4 - x} \right)}}{{4x\left( {2 + \sqrt {4 - x} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{4x\left( {2 + \sqrt {4 - x} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{1}{{4\left( {2 + \sqrt {4 - x} } \right)}} = \frac{1}{{4.\left( {2 + \sqrt {4 - 0} } \right)}} = \frac{1}{{16}}
\end{array}\)

Câu 2:

Hàm số đã cho có đạo hàm tại \(x = 2\) nên ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} {x^2} = {2^2} = 4\\
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( { - \frac{{{x^2}}}{2} + bx - 6} \right) = \frac{{ - {2^2}}}{2} + b.2 - 6 = 2b - 8\\
 \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \Leftrightarrow 2b - 8 = 4 \Leftrightarrow b = 6\\
3,\\
y = \frac{{ - {x^2} + 2x - 3}}{{x - 2}} = \frac{{ - x\left( {x - 2} \right) - 3}}{{x - 2}} =  - x - \frac{3}{{x - 2}}\\
 \Rightarrow y' =  - 1 + \frac{3}{{{{\left( {x - 2} \right)}^2}}}
\end{array}\)