18) Its nothing fancy when it comes to series-- its a geometric series, as @TheYoungin points out.
So its probably easier to sum and then take the derivative. Sometimes its easier to do the term-by-term derivative and then to sum the series, but not in this case.
f(x) = 1/(1+x^2)
f'(x) = -2x/(1+x^2)^2
19) Let D be the derivative, considered as an operator, and then factor the equation they way you would a polynomial
D^3 y - 3D^2 y +3 Dy - y =0
or (D^3 - 3D^2 + 3D - 1)y = 0
or (D-1)^3 y = 0
So one solution is given by solving (D-1)y = 0, which is yields y=e^t (in general, a factor of (D-r) gives a solution of e^rt )
So given a solution of e^t and a repeated factor, the next is given by solving (D-1)y = e^t, which happens to a have a solution te^t.
Given a solution of the form te^t and a factor which appears three times, the next
is given by solving (D-1)y = te^t, which happens to have a solution 1/2 t^2 e^t
In general, if you have k+1 repeated factors like (D-r) in the equation, then you can iteratively solve starting with e^rt to get a general solution like 1/k! t^k e^rt (though the leading constant factor doesn't really matter). You can see this taking the solution e^rt and using the technique of variation of parameters to try a solution of the form y=u(t) e^rt to get an equation like u'(t) = t^(k-1)/(k-1)!, which you can directly integrate.