Induction proof for the n-th derivative of arctangent

Key takeaway: This proof is really about spotting the pattern, then showing it survives one more derivative. The polynomial part is there to absorb the messy algebra so the shape of the result stays clean.

We prove by mathematical induction that for all positive integers $n$, $\frac{d^n}{dx^n}(\tan^{-1}x)=P_n(x)(1+x^2)^{-n},$ where $P_n(x)$ is a polynomial of degree $n-1$.

Base case: $n=1$

We know $\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}=(1+x^2)^{-1}.$ So the result holds with $P_1(x)=1,$ which is a polynomial of degree $0=n-1$.

Inductive hypothesis

Assume that for some $n\ge 1$, $\frac{d^n}{dx^n}(\tan^{-1}x)=P_n(x)(1+x^2)^{-n},$ where $P_n(x)$ is a polynomial of degree $n-1$.

Inductive step

Differentiate both sides:

=\frac{d}{dx}\left[P_n(x)(1+x^2)^{-n}\right].$$ Using the product rule, $$=P_n'(x)(1+x^2)^{-n}+P_n(x)\cdot(-n)(1+x^2)^{-n-1}(2x).$$ Factor out $(1+x^2)^{-(n+1)}$: $$=(1+x^2)^{-(n+1)}\left[P_n'(x)(1+x^2)-2nxP_n(x)\right].$$ Define $$P_{n+1}(x)=P_n'(x)(1+x^2)-2nxP_n(x).$$ Then $$\frac{d^{n+1}}{dx^{n+1}}(\tan^{-1}x)=P_{n+1}(x)(1+x^2)^{-(n+1)}.$$ Now check the degree of $P_{n+1}$. Since $P_n$ has degree $n-1$: - $P_n'$ has degree $n-2$, - multiplying by $(1+x^2)$ gives degree at most $n$, - $xP_n(x)$ also has degree $n$. So $P_{n+1}(x)$ is a polynomial of degree $n$. Therefore the statement holds for $n+1$. By induction, the result is true for all positive integers $n$. --- **Pitfalls the pros know** 👇 The easiest place to lose the thread is the degree count. After differentiating, the polynomial is not just whatever comes out of the product rule; you still need to justify that its degree is exactly one more than before. Another common mistake is forgetting to factor out $(1+x^2)^{-(n+1)}$ before naming the new polynomial. **What if the problem changes?** If you wanted a similar proof for $\ln(1+x^2)$ or $\arctan(ax)$, the structure is almost the same: write the $n$th derivative in a factored form, differentiate once more, and package the algebra into a new polynomial. For $\arctan(ax)$, an extra constant $a^n$ would appear in the derivative pattern because each differentiation brings down another factor of $a$ through the chain rule. `Tags`: mathematical induction, higher derivatives, polynomial degree