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Question

# fn(x)=efn−1(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{fn−1(x)} B fn(x)fn−1(x)⋯f2(x).f1(x)ddx{fn(x)}=ddx{efn−1(x)}⇒efn−1(x)ddx{fn−1(x)}=fn(x)ddx{fn−1(x)}⇒fn(x).ddx{efn−2(x)}=fn(x).efn−2(x)ddx{fn−2(x)}⇒fn(x)fn−1(x)ddx{fn−2(x)}⋯⇒fn(x)fn−1(x)⋯f2(x)ddx{f1(x)}⇒fn(x)fn−1(x)⋯f2(x)ddx{ef0(x)}⇒fn(x)fn−1(x)⋯f2(x)ef0(x)ddx{f0(x)}Use ef0(x)=f1(x) and f0(x)=xHence, ddx{fn(x)}=fn(x)fn−1(x)...f2(x)f1(x)Hence, option C is correct.

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