Let $f(x) = x-\tfrac1{x}$, and define $f^1(x) = f(x)$ and $f^n(x) = f(f^{n-1}(x))$ for $n\ge2$. For each $n$, there is a minimal degree $d_n$ such that there exist polynomials $p$ and $q$ with $f^n(x) = \tfrac{p(x)}{q(x)}$ and the degree of $q$ is equal to $d_n$. Find $d_n$.
1. **Define the function and its iterations:**
   Let \( f(x) = x - \frac{1}{x} \). Define \( f^1(x) = f(x) \) and for \( n \geq 2 \), define \( f^n(x) = f(f^{n-1}(x)) \).

2. **Establish the recurrence relations:**
   Consider the sequences \( S(n) \) and \( T(n) \) defined as follows:
   \[
   S(1) = x, \quad T(1) = 1
   \]
   For \( n > 1 \):
   \[
   S(n) = S(n-1)^2 + T(n-1)^2, \quad T(n) = S(n-1)T(n-1)
   \]

3. **Express \( f^n(x) \) in terms of \( S(n) \) and \( T(n) \):**
   We aim to show that:
   \[
   f^n(x) = \frac{S(n+1)}{T(n+1)}
   \]
   **Base case:**
   For \( n = 1 \):
   \[
   f^1(x) = x - \frac{1}{x} = \frac{x^2 - 1}{x} = \frac{S(2)}{T(2)}
   \]
   where \( S(2) = x^2 - 1 \) and \( T(2) = x \).

4. **Inductive step:**
   Assume the lemma holds for some \( k \geq 1 \). Then:
   \[
   f^k(x) = \frac{S(k+1)}{T(k+1)}
   \]
   We need to show that:
   \[
   f^{k+1}(x) = f(f^k(x)) = \frac{S(k+2)}{T(k+2)}
   \]
   Using the recurrence relations:
   \[
   f^{k+1}(x) = f\left( \frac{S(k+1)}{T(k+1)} \right) = \frac{S(k+1)}{T(k+1)} - \frac{T(k+1)}{S(k+1)}
   \]
   Simplifying:
   \[
   f^{k+1}(x) = \frac{S(k+1)^2 - T(k+1)^2}{S(k+1)T(k+1)} = \frac{S(k+2)}{T(k+2)}
   \]
   where \( S(k+2) = S(k+1)^2 - T(k+1)^2 \) and \( T(k+2) = S(k+1)T(k+1) \).

5. **Irreducibility of the fraction:**
   We need to show that \( \frac{S(n)^2 - T(n)^2}{S(n)T(n)} \) is irreducible. This is equivalent to showing that \( \gcd(S(n), T(n)) = 1 \).

   **Base case:**
   For \( n = 1 \):
   \[
   \gcd(x, 1) = 1
   \]

6. **Inductive step:**
   Assume \( \gcd(S(k), T(k)) = 1 \). We need to show that:
   \[
   \gcd(S(k+1), T(k+1)) = \gcd(S(k)^2 - T(k)^2, S(k)T(k)) = 1
   \]
   Let \( d = \gcd(S(k)^2 - T(k)^2, S(k)T(k)) \). If \( d \neq 1 \), let \( p \) be a prime such that \( p \mid d \). Since \( p \mid S(k)T(k) \) and \( \gcd(S(k), T(k)) = 1 \), \( p \) must divide exactly one of \( S(k) \) or \( T(k) \). However, since \( p \mid S(k)^2 - T(k)^2 \), if \( p \mid S(k) \), then \( p \mid T(k) \), which is a contradiction. Thus, \( d = 1 \).

7. **Degree of the polynomials:**
   Let \( f^n(x) = \frac{p_n(x)}{q_n(x)} \). From the above, \( p_n(x) \) and \( q_n(x) \) are minimal degree polynomials. We claim that:
   \[
   \frac{\deg p_n}{\deg q_n} = \frac{2^n}{2^n - 1}
   \]

8. **Inductive proof of the degree ratio:**
   Assume the claim holds for some \( k \). Then:
   \[
   f^{k+1}(x) = f^k(x) - \frac{1}{f^k(x)} = \frac{p_k}{q_k} - \frac{q_k}{p_k} = \frac{p_k^2 - q_k^2}{p_k q_k}
   \]
   Substituting \( p_k = x^{2^n} \) and \( q_k = x^{2^n - 1} \):
   \[
   \frac{(x^{2^n})^2 - (x^{2^n - 1})^2}{(x^{2^n})(x^{2^n - 1})} = \frac{x^{2^{n+1}} - x^{2^{n+1} - 2}}{x^{2^{n+1} - 1}}
   \]
   Thus, the induction is complete.

9. **Base case for degree:**
   For \( k = 1 \):
   \[
   f^1(x) = \frac{x^2 - 1}{x} \quad \text{and} \quad \deg(x) = 2^1 - 1 = 1
   \]

10. **Conclusion:**
    Therefore, \( d_n = 2^n - 1 \).

The final answer is \( \boxed{ d_n = 2^n - 1 } \)