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fn(x)=e^(f(n-1)(x)) for all n in Na n d...

`f_n(x)=e^(f_(n-1)(x))` for all `n in Na n df_0(x)=x ,t h e n d/(dx){f_n(x)}` is `(f_n(x)d)/(dx){f_(n-1)(x)}` (b) `f_n(x)f_(n-1)(x)` `f_n(x)f_(n-1)(x)f_2(x)dotf_1(x)` `non eoft h e s e`

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

The correct Answer is:
A, C, D
`d/(dx)f_(n)(x)=d/(dx){e^(f_(n-1)(x))}`
`e^(f_(n-1)(x)) d/(dx)f_(n-1)(x)`
`impliesf_(n)(x) d/(dx)(f_(n-1)(x))`
`=f_(n)(x)d/(dx) (e^(f_(n-2)(x)))`
`impliesf_(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)`
`impliesf_(n)(x)f_(n-1)(x)...f_(2)(x) d/(dx)f_(1)(x)`
`=f_(n)(x)f_(n-1)(x)......f_(2)(x)e^(f_(@)(X))d/(dx)f_(@)(x)`
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