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(B) Given that $f^{\prime}(x)=f(x)$ for all $x \in \mathbb{R}$.This is a first-order linear differential equation: $\frac{f^{\prime}(x)}{f(x)}=1$.Inte

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$4 e$

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(B) Given that $f^{\prime}(x)=f(x)$ for all $x \in \mathbb{R}$.This is a first-order linear differential equation: $\frac{f^{\prime}(x)}{f(x)}=1$.Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$.Thus,$f(x) = e^{x+C} = k e^x$,where $k = e^C$ is a constant.Given $f(1)=2$,we have $k e^1 = 2$,which implies $k = \frac{2}{e}$.So,$f(x) = \frac{2}{e} \cdot e^x = 2 e^{x-1}$.Now,$h(x) = f(f(x))$.Using the chain rule,$h^{\prime}(x) = f^{\prime}(f(x)) \cdot f^{\prime}(x)$.Since $f^{\prime}(x) = f(x)$,we have $h^{\prime}(x) = f(f(x)) \cdot f(x)$.At $x=1$,$h^{\prime}(1) = f(f(1)) \cdot f(1)$.Since $f(1)=2$,$h^{\prime}(1) = f(2) \cdot 2$.Using $f(x) = 2 e^{x-1}$,we find $f(2) = 2 e^{2-1} = 2e$.Therefore,$h^{\prime}(1) = (2e) \cdot 2 = 4e$.

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

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