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Question
Let f1(x)=ex,f2(x)=ef1(x),.....fn+1(x)=efn(x) for all n≥1. The for any fixed n,ddxfn(x) is
Let f1(x)=ex,f2(x)=ef1(x),.....fn+1(x)=efn(x) for all n≥1. The for any fixed n,ddxfn(x) is
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Solution
The correct option is C fn(x)fn−1(x)....f1(x)
ddxfr(x)=ddxefr−1(x)=efr−1(x)ddxfr−1(x)=fr(x)ddxfr−1(x) ∀r∈N>1
∴ddxfn(x)=fn(x)ddxfn−1(x)=fn(x)fn−1(x)ddxfn−2(x)=fn(x)fn−1(x)fn−2(x)ddxfn−3(x)...=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)ddxf1(x)
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)ddxex
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)ex
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)f1(x)
Hence, the answer is option (c).
ddxfr(x)
=ddxefr−1(x)
=efr−1(x)ddxfr−1(x)
=fr(x)ddxfr−1(x) ∀r∈N>1
∴ddxfn(x)=fn(x)ddxfn−1(x)
=fn(x)fn−1(x)ddxfn−2(x)
=fn(x)fn−1(x)fn−2(x)ddxfn−3(x)
.
.
.
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)ddxf1(x)
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)ddxex
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)ex
=fn(x)fn−1(x)fn−2(x)....f3(x)f2(x)f1(x)
Hence, the answer is option (c).
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