calculus – $alpha, beta$ are roots of $x^2+px+q=0, f(x)=(alpha+beta)x-left(frac{alpha^2+beta^2}2right)x^2+left(frac{alpha^3+beta^3}3right)x^3-…$

If $ln(1+x)=x-frac {x^2}2+frac {x^3}3-frac{x^4}4+…$ and $alpha$ & $beta$ are roots of the equation $x^2+px+q=0$. If $f(x)=(alpha+beta)x-left(frac{alpha^2+beta^2}2right)x^2+left(frac{alpha^3+beta^3}3right)x^3-left(frac{alpha^4+beta^4}2right)x^4+…$ then find $f'(x)$ in terms of $p,q$

$alpha+beta=-p,; alphabeta=q$

$ln(1+(alpha+beta)x)=(alpha+beta)x-frac {(alpha+beta)^2}2x^2+frac {(alpha+beta)^3}3x^3-frac{(alpha+beta)^4}4x^4+…$

$f'(x)=(alpha+beta)-(alpha^2+beta^2)x+(alpha^3+beta^3)x^2-(alpha^4+beta^4)x^3$

$alpha^2+beta^2=(alpha+beta)^2-2alphabeta=p^2-2q$

$alpha^3+beta^3=(alpha+beta)^3-3alphabeta(alpha+beta)=-p^3+3pq$

Not able to work around and conclude.