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Question

If fn(x)=efn1(x) for all nϵN and f0(x)=x then ddx{fn(x)} is equal to

A
fn(x).ddx{fn1(x)}
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B
fn(x).fn1(x)
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C
fn(x).fn1(x)....f2(x).f1(x)
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D
none of these
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Solution

The correct options are
A fn(x).ddx{fn1(x)}
C fn(x).fn1(x)....f2(x).f1(x)
Given: fn(x)=efn1(x)
Consider, ddx{fn(x)}
=ddxefn1(x)
=efn1(x)ddx(fn1(x))
=fn(x)ddx(fn1(x))
=fn(x)ddxefn2(x)
=fn(x)efn2(x)ddxfn2(x)
=fn(x)fn1(x)ddxfn2(x)
Continuing this process, we get
ddx{fn(x)}
=fn(x)fn1(x)fn2(x)....f1(x)ddxf0(x)

Hence, ddx{fn(x)}=fn(x)fn1(x)fn2(x)....f1(x)

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