`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`

Options

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

Answer

A, C, D

Explanation

`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)`