DifferentiationHard
Question
If fn (x) = efn-1(x) for all n ∈ N and fo (x) = x, then 
{fn (x)} is equal to
{fn (x)} is equal toOptions
A.fn (x) . 
{fn - 1 (x)}
{fn - 1 (x)}B.fn (x). fn - 1 (x)
C.fn (x). fn - 1 (x)........ f2 (x). f1 (x)
D.none of these
Solution
fn(x) = efn-1(x) ∀ n ∈ N and f0 (x) = x
f1(x) = ex
f2(x) = eex
f3(x) = eeex
:
:
fn(x) = eee.......n times x = efn-1(x)
Now
fn(x) = fn(x) 
= fn(x) .fn-1 (x)
:
:
= fn(x).fn-1 (x).fn-2 (x) ........f1(x).1
f1(x) = ex
f2(x) = eex
f3(x) = eeex
:
:
fn(x) = eee.......n times x = efn-1(x)
Now
fn(x) = fn(x) = fn(x) .fn-1 (x)
:
:
= fn(x).fn-1 (x).fn-2 (x) ........f1(x).1
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