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Question

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

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)}
B fn(x)fn1(x)f2(x).f1(x)
ddx{fn(x)}=ddx{efn1(x)}
efn1(x)ddx{fn1(x)}=fn(x)ddx{fn1(x)}
fn(x).ddx{efn2(x)}=fn(x).efn2(x)ddx{fn2(x)}
fn(x)fn1(x)ddx{fn2(x)}

fn(x)fn1(x)f2(x)ddx{f1(x)}
fn(x)fn1(x)f2(x)ddx{ef0(x)}
fn(x)fn1(x)f2(x)ef0(x)ddx{f0(x)}
Use ef0(x)=f1(x) and f0(x)=x
Hence,
ddx{fn(x)}=fn(x)fn1(x)...f2(x)f1(x)
Hence, option C is correct.

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