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

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

which is true.

Inductive step: Assume

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

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

Then,

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

This can be written as

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

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

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