Question

$f_n\left(x\right)=e^{f_{n-1}\left(x\right)}$ for all $nϵN$ and $f_0\left(x\right)=x$, then $\frac{d}{dx}\left\{f_n\right(x\left)\right\}$ is

• A. $f_n\left(x\right)\frac{d}{dx}\left\{f_{n-1}\right(x\left)\right\}$
• B. $f_n\left(x\right)f_{n-1}\left(x\right)$
• C. $f_n\left(x\right)f_{n-1}\left(x\right)\cdots f_2\left(x\right).f_1\left(x\right)$
• D. none of these

Answer 1

Answer: (A), (C)

$f_n\left(x\right)\frac{d}{dx}\left\{f_{n-1}\right(x\left)\right\}$
$f_n\left(x\right)f_{n-1}\left(x\right)\cdots f_2\left(x\right).f_1\left(x\right)$

$\frac{d}{dx}\left\{f_n\right(x\left)\right\}=\frac{d}{dx}\left\{e^{f_{n-1}\left(x\right)}\right\}$
$\Rightarrow e^{f_{n-1}\left(x\right)}\frac{d}{dx}\left\{f_{n-1}\right(x\left)\right\}=f_n\left(x\right)\frac{d}{dx}\left\{f_{n-1}\right(x\left)\right\}$
$\Rightarrow f_n\left(x\right).\frac{d}{dx}\left\{e^{f_{n-2}\left(x\right)}\right\}=f_n\left(x\right).e^{f_{n-2}\left(x\right)}\frac{d}{dx}\left\{f_{n-2}\right(x\left)\right\}$
$\Rightarrow f_n\left(x\right)f_{n-1}\left(x\right)\frac{d}{dx}\left\{f_{n-2}\right(x\left)\right\}$
$\cdots$
$\Rightarrow f_n\left(x\right)f_{n-1}\left(x\right)\cdots f_2\left(x\right)\frac{d}{dx}\left\{f_1\right(x\left)\right\}$
$\Rightarrow f_n\left(x\right)f_{n-1}\left(x\right)\cdots f_2\left(x\right)\frac{d}{dx}\left\{e^{f_0\left(x\right)}\right\}$
$\Rightarrow f_n\left(x\right)f_{n-1}\left(x\right)\cdots f_2\left(x\right)e^{f_0\left(x\right)}\frac{d}{dx}\left\{f_0\right(x\left)\right\}$
Use $e^{f_0\left(x\right)}=f_1\left(x\right)$ and $f_0\left(x\right)=x$
Hence,
$\frac{d}{dx}\left\{f_n\right(x\left)\right\}=f_n\left(x\right)f_{n-1}\left(x\right)...f_2\left(x\right)f_1\left(x\right)$
Hence, option C is correct.