Let $E$ be the set of all continuously differentiable real valued functions $f$ on $[0,1]$ such that $f(0)=0$ and $f(1)=1$. Define
$$J(f)=\int^1_0(1+x^2)f'(x)^2\text dx.$$
a) Show that $J$ achieves its minimum value at some element of $E$.
b) Calculate $\min_{f\in E}J(f)$.
1. **Show that \( J \) achieves its minimum value at some element of \( E \):**

   We start by considering the functional 
   \[
   J(f) = \int_0^1 (1 + x^2) f'(x)^2 \, dx
   \]
   for \( f \in E \), where \( E \) is the set of all continuously differentiable real-valued functions on \([0,1]\) such that \( f(0) = 0 \) and \( f(1) = 1 \).

   To show that \( J \) achieves its minimum value, we use the Cauchy-Schwarz inequality. Let \( g(x) = \sqrt{1 + x^2} f'(x) \) and \( h(x) = \frac{1}{\sqrt{1 + x^2}} \). Then, by the Cauchy-Schwarz inequality, we have:
   \[
   \left( \int_0^1 g(x) h(x) \, dx \right)^2 \leq \left( \int_0^1 g(x)^2 \, dx \right) \left( \int_0^1 h(x)^2 \, dx \right)
   \]
   Substituting \( g(x) \) and \( h(x) \), we get:
   \[
   \left( \int_0^1 f'(x) \, dx \right)^2 \leq \left( \int_0^1 (1 + x^2) f'(x)^2 \, dx \right) \left( \int_0^1 \frac{1}{1 + x^2} \, dx \right)
   \]
   Since \( f(1) - f(0) = 1 \), we have:
   \[
   1^2 \leq J(f) \cdot \int_0^1 \frac{1}{1 + x^2} \, dx
   \]
   The integral \( \int_0^1 \frac{1}{1 + x^2} \, dx \) is the arctangent function evaluated from 0 to 1:
   \[
   \int_0^1 \frac{1}{1 + x^2} \, dx = \arctan(1) - \arctan(0) = \frac{\pi}{4}
   \]
   Therefore, we have:
   \[
   1 \leq J(f) \cdot \frac{\pi}{4}
   \]
   which implies:
   \[
   J(f) \geq \frac{4}{\pi}
   \]
   Equality holds if and only if \( g(x) = \lambda h(x) \) for some constant \( \lambda \). This means:
   \[
   \sqrt{1 + x^2} f'(x) = \frac{\lambda}{\sqrt{1 + x^2}}
   \]
   Solving for \( f'(x) \), we get:
   \[
   f'(x) = \frac{\lambda}{1 + x^2}
   \]
   Integrating both sides, we find:
   \[
   f(x) = \lambda \int_0^x \frac{1}{1 + t^2} \, dt = \lambda \arctan(x)
   \]
   Given the boundary conditions \( f(0) = 0 \) and \( f(1) = 1 \), we have:
   \[
   1 = \lambda \arctan(1) = \lambda \cdot \frac{\pi}{4}
   \]
   Solving for \( \lambda \), we get:
   \[
   \lambda = \frac{4}{\pi}
   \]
   Therefore, the function that minimizes \( J(f) \) is:
   \[
   f(x) = \frac{4}{\pi} \arctan(x)
   \]

2. **Calculate \( \min_{f \in E} J(f) \):**

   Substituting \( f(x) = \frac{4}{\pi} \arctan(x) \) into \( J(f) \), we get:
   \[
   J(f) = \int_0^1 (1 + x^2) \left( \frac{d}{dx} \left( \frac{4}{\pi} \arctan(x) \right) \right)^2 \, dx
   \]
   Since \( \frac{d}{dx} \left( \frac{4}{\pi} \arctan(x) \right) = \frac{4}{\pi} \cdot \frac{1}{1 + x^2} \), we have:
   \[
   J(f) = \int_0^1 (1 + x^2) \left( \frac{4}{\pi} \cdot \frac{1}{1 + x^2} \right)^2 \, dx = \int_0^1 \left( \frac{4}{\pi} \right)^2 \cdot \frac{1}{1 + x^2} \, dx
   \]
   Simplifying, we get:
   \[
   J(f) = \left( \frac{4}{\pi} \right)^2 \int_0^1 \frac{1}{1 + x^2} \, dx = \left( \frac{4}{\pi} \right)^2 \cdot \frac{\pi}{4} = \frac{16}{\pi^2} \cdot \frac{\pi}{4} = \frac{4}{\pi}
   \]

The final answer is \( \boxed{ \frac{4}{\pi} } \)