Example 8 Find the smallest real number $A$, such that for every quadratic trinomial $f(x)$ satisfying the condition $|f(x)| \leqslant 1(0 \leqslant x \leqslant 1)$, the inequality $f^{\prime}(0) \leqslant A$ holds.
Let the quadratic trinomial be \( f(x) = ax^2 + bx + c \). When \( 0 \leq x \leq 1 \), it satisfies the inequality \( |f(x)| \leq 1 \). Therefore,
\[ 
|f(0)| \leq 1, \left|f\left(\frac{1}{2}\right)\right| \leq 1, |f(1)| \leq 1.
\]

Since \( f(0) = c, f\left(\frac{1}{2}\right) = \frac{a}{4} + \frac{b}{2} + c, f(1) = a + b + c, f'(0) = b = 4\left(\frac{a}{4} + \frac{b}{2} + c\right) - (a + b + c) - 3c \),
we have \( |f'(0)| \leq 4\left|f\left(\frac{1}{2}\right)\right| + |f(1)| + 3|f(0)| \leq 4 + 1 + 3 = 8 \). Thus, \( A \leq 8 \).

On the other hand, the quadratic trinomial \( f(x) = -8x^2 + 8x - 1 \) satisfies \( |f(x)| \leq 1 \) (when \( x \in [0,1] \)), and \( f'(x) = -16x + 8, f'(0) = 8 \), so \( A \geq 8 \).

In summary, \( A = 8 \).