Let f be a differentiable function such that f(1) = 2 and f ^ ' (x)...

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

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

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

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

$⇒ \dfrac{{f}^{'}\left(x\right)}{f\left(x\right)} = 1$

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

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

$⇒ f\left(x\right) = k . {e}^{x}$

$⇒ f\left(x\right) = \dfrac{2}{e} . {e}^{x} . . . \left(2\right)\left[∵ f\left(1\right) = 2\right]$

Putting $x = 1$ in equation \left(1\right)$, we get</p><p>${h}^{'}\left(1\right) = f\left(f\left(1\right)\right) . f\left(1\right) = f\left(2\right) . 2 = 2 . \dfrac{2}{e}{e}^{2} = 4e$(using equation \left(2\right)$)