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Question
If fn(x)=efn−1(x) for all nϵN and f0(x)=x then ddx{fn(x)} is equal to
If fn(x)=efn−1(x) for all nϵN and f0(x)=x then ddx{fn(x)} is equal to
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Solution
The correct options are
A fn(x).ddx{fn−1(x)}
C fn(x).fn−1(x)....f2(x).f1(x)
Given: fn(x)=efn−1(x) Consider, ddx{fn(x)}=ddxefn−1(x)=efn−1(x)ddx(fn−1(x))=fn(x)ddx(fn−1(x))=fn(x)ddxefn−2(x)=fn(x)efn−2(x)ddxfn−2(x)=fn(x)fn−1(x)ddxfn−2(x)Continuing this process, we getddx{fn(x)}
=fn(x)fn−1(x)fn−2(x)....f1(x)ddxf0(x)
Hence, ddx{fn(x)}=fn(x)fn−1(x)fn−2(x)....f1(x)
A fn(x).ddx{fn−1(x)}
C fn(x).fn−1(x)....f2(x).f1(x)
Given: fn(x)=efn−1(x)
Consider, ddx{fn(x)}
=ddxefn−1(x)
=efn−1(x)ddx(fn−1(x))
=fn(x)ddx(fn−1(x))
=fn(x)ddxefn−2(x)
=fn(x)efn−2(x)ddxfn−2(x)
=fn(x)fn−1(x)ddxfn−2(x)
Continuing this process, we get
ddx{fn(x)}
=fn(x)fn−1(x)fn−2(x)....f1(x)ddxf0(x)
Hence, ddx{fn(x)}=fn(x)fn−1(x)fn−2(x)....f1(x)
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