Let \(\mathrm{f}\) be a differentiable function such that \(\mathrm{f}(1)=2\) and \(\mathrm{f}^{\prime}(x)=\mathrm{f}(x)\), for all \(x \in \mathrm{R}\). If \(\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))\), then \(\mathrm{h}^{\prime}(1)\) is equal to

Given: \(\mathrm{f}^{\prime}(x)=\mathrm{f}(x)\) for all \(x \in \mathrm{R}^{\prime}\)
\(\Rightarrow \frac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}=1\)
Integrating on both sides, wwe get \(\log |\mathrm{f}(x)|=x+\mathrm{c}\)
\(\Rightarrow \mathrm{f}(x)=\mathrm{e}^{x+c}\)
\(\Rightarrow \mathrm{f}(x)=\mathrm{e}^x \cdot \mathrm{e}^{\mathrm{c}}\)
\(\Rightarrow \mathrm{f}(x)=\mathrm{e}^x \cdot \mathrm{c}_1\)... (i)\(\left[\mathrm{e}^{\mathrm{c}}=\mathrm{c}_1\right]\)
As \(f(1)=2\)
\(\begin{aligned}
& \therefore \quad \mathrm{c}_1 \cdot \mathrm{e}=2 \\
& \Rightarrow \mathrm{c}_1=\frac{2}{\mathrm{e}}
\end{aligned}\)
Equation (i) becomes
\(\mathrm{f}(x)=\mathrm{e}^x \cdot \frac{2}{\mathrm{e}}\)
Now, \(\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))\)
\(\begin{array}{ll}
\therefore \quad \mathrm{h}^{\prime}(x)=\mathrm{f}^{\prime}(\mathrm{f}(x)) \times \mathrm{f}^{\prime}(x) \\
\therefore \quad \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(\mathrm{f}(1)) \times \mathrm{f}^{\prime}(1) \\
\Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(2) \times \mathrm{f}^{\prime}(1) \\
\Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{e}^2 \times \frac{2}{\mathrm{e}} \times 2 \\
\Rightarrow \mathrm{h}^{\prime}(1)=4 \mathrm{e}
\end{array}\)