I basically used the formula that d/dx [f(x)^g(x)] equals:

( f(x)^g(x) ) ( g'(x) ln(f(x)) g(x)f'(x)/f(x) )

& I basically set f(x) = x & g(x) = x^a-1 because f(x)^g(x) = x^a to get the derivative of x^a .

I repeated this process 2 more times & also created the notation (x|a) for d/dx [x^a] & (x||a) for d²/dx² [x^a] (Ik that because of how simple these notations are, they're probably uswd for something else but idc)

So is there a formula to get the nth derivative like the ones below?

dⁿ/dxⁿ a^x = a^x ln(a)ⁿ (this one also works for integrals if n<0)

dⁿ/dxⁿ 1/x = -n! / (-x)ⁿ⁺¹ (this one does not work for integrals since factorials don't work for negative integers)