<s>[INST] 40th derivative of a function I would like to have some verification to see if my answer is correct. The given function is $f(x)=ln(1+x^2)$ and I need the 40th derivative at $x=0$. Here is my work: Using series one can manipulate $\frac{1}{1+x}=1-x+x^2-x^3+x^4...=SUMx^n(-1)^n$ into $\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8...=SUMx^{2n}(-1)^n$. Then $\frac{2x}{1+x}=2x-2x^3+2x^5-2x^7+2x^9...=2SUMx^{2n+1}(-1)^n$. Integrating gives $ln(1+x^2)=...=2SUM\frac{x^{2n+2}(-1)^n}{2n+2}$ where a $2$ cancels to arrive at a $n+1$ in that denominator. Now for the 40th derivative, $2n+2=40$ gives $n=19$ and thus I believe the answer is $\frac{-40!}{20}$ Do you concur? If not could you correct me? Thanks.
 [/INST] Taking the $40$th derivative at 0 will yield you the coefficient for the $x^{40}$ term multiplied by $40!$
The general series for natural log is:
$$
ln(x+1) = \frac {x^1}1 - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4...
$$
replacing $x^2$ makes:
$$
ln(x^2+1) = \frac {x^2}1 - \frac {x^4}2 + \frac {x^6}3 - \frac {x^8}4...
$$
From this it is apparent that the 20th term of the first series shares its coefficient with the 40th term of the second series.
The coefficient equals $-\frac 1{20}$. Multiplying $40!$ to cancel the implied division yields: $-\frac {40!}{20}$
From this I confirm your answer to be correct; however, your proof and solution is incredibly messy. There is a known series for natural log. Using a fractional series and integrating was a waste of time and effort.
</s>