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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_(n)(x)(d)/(dx){f_(n-1)(x)}`

B

`f_(n)(x)f_(n-1)(x)`

C

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

D

None of these

Text Solution

Verified by Experts

`(d)/(dx){f_(n)(x)}=(d)/(dx){e^(f_(n-1)(x))}`
`=e^(f_(n-1)(x))(d)/(dx){f_(n-1)(x)}=f_(n)(x)(d)/(dx){f_(n-1)(x)}`
`f_(n)(x).(d)/(dx){e^(f_(n-2)(x))}=f_(n)(x).e^(f_(n-2)(x))(d)/(dx){f_(n-2)(x)}`
`=f_(n)(x)f_(n-1)(x)(d)/(dx){f_(n-2)(x)}` ...
`=f_(n)(x)f_(n-1)(x)...f_(2)(x)(d)/(dx){f_(1)(x)}`
`=f_(n)(x)f_(n-1)(x)...f_(2)(x)(d)/(dx){e^(f_(0)(x))}`
`=f_(n)(x)f_(n-1)(x)...f_(2)(x)e^(f_(0)(x))(d)/(dx){f_(0)(x)}`
`"Use "e^(f_(0)(x))=f_(1)(x)and f_(0)(x)=x`
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