Let $f$ be a differentiable function such that $f\left(1\right)=2$ and $f^{\text{'}}\left(x\right)=f\left(x\right)$ for all $x\in R$. If $h\left(x\right)=f\left(f\left(x\right)\right)$, then $h^{\text{'}}\left(1\right)$ is equal to :

Given $h\left(x\right)=f\left(f\left(x\right)\right)$

$\Rightarrow h^{\text{'}}\left(x\right)=f^{\text{'}}\left(f\left(x\right)\right).f^{\text{'}}\left(x\right)$

$\Rightarrow h^{\text{'}}\left(x\right)=f\left(f\left(x\right)\right).f\left(x\right) ...\left(1\right)$ (as $f^{\text{'}}\left(x\right)=f\left(x\right)$) 

Now $f^{\text{'}}\left(x\right)=f\left(x\right)$

$\Rightarrow \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=1$

Integrating both sides with respect to $x$, we get

$\ln \left|f\left(x\right)\right|=x+c$

$\Rightarrow f\left(x\right)=k.e^x$

$\Rightarrow f\left(x\right)=\frac{2}{e}. e^x ...\left(2\right) \left[\because f\left(1\right)=2\right]$

Putting $x=1$ in equation $\left(1\right)$, we get

$h^{\text{'}}\left(1\right)=f\left(f\left(1\right)\right).f\left(1\right)=f\left(2\right).2=2.\frac{2}{e}{e}^2=4e$ (using equation $\left(2\right)$)