Let f be a differentiable function such that

Solution:

Since, $\quad f^{\prime}(x)=f(x)$

Then, $\quad \frac{f^{\prime}(x)}{f(x)}=1$

$\Rightarrow \frac{f^{\prime}(x)}{f(x)}=d x \Rightarrow \frac{f^{\prime}(x)}{f(x)} \quad d x=\int d x$

$\Rightarrow \quad \ln |f(x)|=x+c$

$f(x)=\pm e^{x+c}$......(1)

Since, the given condition

$f(1)=2$

From eq $^{\mathrm{n}}(1) \quad f(x)=e^{x+c}=e^{c} e^{x}$

Then, $f(1)=e^{c} \cdot e^{1}$

$\Rightarrow \quad 2=e^{c} \cdot e$

$\Rightarrow \quad \frac{2}{e}=e^{c}$

Then, from eqn (1)

$f(x)=\frac{2}{e} e^{x}$

$\Rightarrow \quad f^{\prime}(x)=\frac{2}{e} e^{x}$

Now $\quad h(x)=f(f(x))$

$\Rightarrow \quad h^{\prime}(x)=f^{\prime}(f(x)) \cdot f^{\prime}(x)$

$h^{\prime}(1)=f^{\prime}(2) \cdot f^{\prime}(1)=\frac{2}{e} e^{2} \cdot \frac{2}{e} \cdot e=4 e$