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f(n)(x)=e^(f(n-1)(x))" for all "n in N a...

`f_(n)(x)=e^(f_(n-1)(x))" for all "n in N and f_(0)(x)=x," then "(d)/(dx){f_(n)(x)}` is

A

`f_(a)(x).(d)/(dx){f_(a-1)(x)}`

B

`f_(a)(x).f_(a-1)(x)`

C

`f_(a)(x).f_(a-1)(x)......f_(2)(x).f_(1)(x)`

D

none of these

Text Solution

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The correct Answer is:
A, C
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