let y=f[g[h{u(g(x))}]]

then dy\ax is?

dy/dx=f'(a).a'(x).....where a =g[h{u(v(x))}]

  similarly u expand it further..

for clear explanation...lets take an example of y=sin(logx)

=>dy/dx= cos(logx).1/x

here say h(g(x)) is in the form of sin(log(x))..

if u have any doubt in understanding it u can reply me bac..