Let f be a differentiable function such that f 1 = 2 and f ' x = f x for all x ∈ R . If h x = f f x ,…

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 $\left(2\right)$)