well general method i can give u a very good idea
here
suppose a function be xm
then the nth derevative will be like this see
y=xm
y`=mxm-1
y``=m*(m-1)xm-2
for the nth derevative
dny/dxn=m*(m-1)(m-2)(m-3)(m-4)............(m-n)xm-n
we multiply and divide by n!
dny/dxn=m!/n!xm-n
dny/dxn=mPn xm-n
this is the method
using this we can calculate for
general formula of nth derivative of (ax+b)-r is
(-1)n (n+r-1)! (a)n / (ax+b)n+r (r-1)!
(as wriiten by prasad)
for such problems we need to find a relation between the derevatives!!
as a function of the number of times differenciated!!
these things dont come in JEE!!
all this successive differenciation and leibnitz theorum
i.e sucessive differenciation of product of two functions!!