Base case: For \(n = 1\),

\(\frac{d}{dx} \left( \arctan x \right) = \frac{1}{1 + x^2},\)

which is true.

Inductive step: Assume

\(\frac{d^k}{dx^k} \left( \arctan x \right) = P_k(x) (1 + x^2)^{-k},\)

where \(\deg P_k(x) = k - 1\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( \arctan x \right) = P_k'(x) (1 + x^2)^{-k} - 2kx P_k(x) (1 + x^2)^{-k-1}.\)

This can be written as

\(\left( P_k'(x) (1 + x^2) - 2kx P_k(x) \right) (1 + x^2)^{-k-1},\)

so \(\deg P_{k+1}(x) = k\).

Thus, true for \(n = k + 1\). By induction, true for all positive integers \(n\).