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Question

If fn1(x)=ln(fn(x)) nN and f0(x)=x1, then ddx(fn(x)) is

A
fn(x)fn1(x)
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B
fn(x)fn1(x)fn2(x)f1(x)
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C
fn1(x)ddx(fn1(x))
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D
fn(x)ddx(fn1(x))
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Solution

The correct option is D fn(x)ddx(fn1(x))
fn1(x)=ln(fn(x))
Diffrentiating
fn1(x)=1fn(x)×fn(x)
fn(x)=fn(x).fn1(x)
Replace n with n1

fn1(x)=fn1(x).fn2(x)
fn(x)=fn(x).fn1(x).fn2(x)......f1(x).f0(x)....(i)
f0(x)=x1 f0(x)=1
From (i)
fn(x)=fn(x).fn1(x).fn2(x).......f1(x)

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