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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Aicharya · Accepted Answer

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)$\left[\mathrm{e}^{\mathrm{c}}=\mathrm{c}_1ight]$ As $f(1)=2$ $\begin{aligned} & 	herefore \quad \mathrm{c}_1 \cdot \mathrm{e}=2 \ & \Rightarrow \mathrm{c}_1=\frac{2}{\mathrm{e}} \end{aligned}$ Equation (i) becomes $\mathrm{f}(x)=\mathrm{e}^x \cdot \frac{2}{\mathrm{e}}$ Now, $\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))$ $\begin{array}{ll} 	herefore \quad \mathrm{h}^{\prime}(x)=\mathrm{f}^{\prime}(\mathrm{f}(x)) 	imes \mathrm{f}^{\prime}(x) \ 	herefore \quad \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(\mathrm{f}(1)) 	imes \mathrm{f}^{\prime}(1) \ \Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{f}^{\prime}(2) 	imes \mathrm{f}^{\prime}(1) \ \Rightarrow \mathrm{h}^{\prime}(1)=\mathrm{e}^2 	imes \frac{2}{\mathrm{e}} 	imes 2 \ \Rightarrow \mathrm{h}^{\prime}(1)=4 \mathrm{e} \end{array}$

Let $\mathrm{f}$ be a differentiable function such that $\mathrm{f}(1)=2$ and…

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