Find the smallest possible $\alpha\in \mathbb{R}$ such that if $P(x)=ax^2+bx+c$ satisfies $|P(x)|\leq1 $ for $x\in [0,1]$ , then we also have $|P'(0)|\leq \alpha$.
To find the smallest possible $\alpha \in \mathbb{R}$ such that if $P(x) = ax^2 + bx + c$ satisfies $|P(x)| \leq 1$ for $x \in [0,1]$, then we also have $|P'(0)| \leq \alpha$, we need to follow these steps:

1. **Express the derivative of \( P(x) \):**
   \[
   P(x) = ax^2 + bx + c \implies P'(x) = 2ax + b
   \]
   Therefore, at \( x = 0 \):
   \[
   P'(0) = b
   \]

2. **Analyze the condition \( |P(x)| \leq 1 \) for \( x \in [0,1] \):**
   We need to ensure that the polynomial \( P(x) = ax^2 + bx + c \) remains within the bounds \([-1, 1]\) for all \( x \) in the interval \([0, 1]\).

3. **Determine the constraints on \( a, b, \) and \( c \):**
   - At \( x = 0 \):
     \[
     |P(0)| = |c| \leq 1 \implies -1 \leq c \leq 1
     \]
   - At \( x = 1 \):
     \[
     |P(1)| = |a + b + c| \leq 1
     \]
   - At \( x = \frac{1}{2} \):
     \[
     |P\left(\frac{1}{2}\right)| = \left|a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + c\right| = \left|\frac{a}{4} + \frac{b}{2} + c\right| \leq 1
     \]

4. **Find the maximum value of \( |b| \) under these constraints:**
   We need to maximize \( |b| \) while ensuring that \( |P(x)| \leq 1 \) for \( x \in [0,1] \).

5. **Construct a polynomial that meets the conditions:**
   Consider the polynomial \( P(x) = -8x^2 + 8x - 1 \):
   - At \( x = 0 \):
     \[
     P(0) = -1 \quad (\text{which is within the bounds})
     \]
   - At \( x = 1 \):
     \[
     P(1) = -8(1)^2 + 8(1) - 1 = -8 + 8 - 1 = -1 \quad (\text{which is within the bounds})
     \]
   - At \( x = \frac{1}{2} \):
     \[
     P\left(\frac{1}{2}\right) = -8\left(\frac{1}{2}\right)^2 + 8\left(\frac{1}{2}\right) - 1 = -8\left(\frac{1}{4}\right) + 4 - 1 = -2 + 4 - 1 = 1 \quad (\text{which is within the bounds})
     \]

6. **Verify the derivative at \( x = 0 \):**
   \[
   P'(x) = -16x + 8 \implies P'(0) = 8
   \]
   This shows that \( |P'(0)| = 8 \).

Therefore, the smallest possible value of \( \alpha \) is indeed 8.

The final answer is \( \boxed{ \alpha = 8 } \)