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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
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$4 e$
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, $\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))$
$\begin{array}{ll}
\therefore \quad \mathrm{h}^{\prime}(x)=\mathrm{f}^{\prime}(\mathrm{f}(x)) \times \mathrm{f}^{\prime}(x) \\
\therefore \quad \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(\mathrm{f}(1)) \times \mathrm{f}^{\prime}(1) \\
\Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(2) \times \mathrm{f}^{\prime}(1) \\
\Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{e}^2 \times \frac{2}{\mathrm{e}} \times 2 \\
\Rightarrow \mathrm{h}^{\prime}(1)=4 \mathrm{e}
\end{array}$
$\Rightarrow \frac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}=1$
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, $\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))$
$\begin{array}{ll}
\therefore \quad \mathrm{h}^{\prime}(x)=\mathrm{f}^{\prime}(\mathrm{f}(x)) \times \mathrm{f}^{\prime}(x) \\
\therefore \quad \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(\mathrm{f}(1)) \times \mathrm{f}^{\prime}(1) \\
\Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(2) \times \mathrm{f}^{\prime}(1) \\
\Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{e}^2 \times \frac{2}{\mathrm{e}} \times 2 \\
\Rightarrow \mathrm{h}^{\prime}(1)=4 \mathrm{e}
\end{array}$
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