Let f be a differentiable function such that

Question:

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

  1. $4 \mathrm{e}$

  2. $4 \mathrm{e}^{2}$

  3. $2 \mathrm{e}$

  4. $2 \mathrm{e}^{2}$


Correct Option: , 3

Solution:

$\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=1 \forall \mathrm{x} \in \mathrm{R}$

Intergrate \& use $f(1)=2$

$\mathrm{f}(\mathrm{x})=2 \mathrm{e}^{\mathrm{x}-1} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{e}^{\mathrm{x}-1}$

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

$h^{\prime}(1)=f^{\prime}(f(1)) f^{\prime}(1)$

$=\mathrm{f}^{\prime}(2) \mathrm{f}^{\prime}(1)$

$=2 \mathrm{e} \cdot 2=4 \mathrm{e}$

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