Let y=tan^-1(x/a), then tan y = x/a.
sec^2(y)dy/dx=1/a so dy/dx=1/(a(sec^2(y))=1/(a(1+tan^2(y))=1/(a(1+(x^2/a^2))=1/(a+x^2/a)=(a+x^2/a)^-1
Write y'=dy/dx for convenience, so y'' is the second derivative and y'=(a+x^2/a)^-1 or sec^2(y)y'=1/a. Let u=v=sec y and w=y', then sec^2(y)dydx=1/a can be expressed uvw=1/a. Therefore we can get the second derivative by differentiating further: d/dx(u.v.w.)=v.w.du/dx+u.w.dv/dx+u.v.dw/dx=0 (differentiation by parts)
2sec(y).y'.sec(y)tan(y)y'+sec^2(y).y''=0
2sec^2(y)tan(y)y'^2+sec^2(y)y''=0, which we can divide through by sec^2(y):
tan(y)y'^2+y''=0 making y''=-tan(y)y'^2=-x/a((a+x^2/a)^-2)
So far, there's no obvious pattern in the derivatives which would suggest the nth derivative. Let's continue by finding the 3rd derivative, using the derivatives of y rather than converting to expressions in x.
y'''=d/dx(-tan(y)y'^2)
To calculate d/dx(y'^2) we can differentiate by parts and the result is 2y'y''. If we differentiate the right side of this equation we get y'''=-sec^2(y)y'-2tan(y)y'y''=-(1+tan^2(y))y'-2tan(y)y'y''=-y'(1+tan^2(y)+2tan(y)y''). Or, since we know that sec^2(y)y'=1/a, we can substitute to simplify the third derivative: y'''=-1/a-2tan(y)y'y''. Also, y''=-tan(y)y'^2 and tan(y)=x/a, so a simpler expression for y'''=-1/a-(2x/a)y'y''.
More to follow...