Question

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

• A. $2e^2$
• B. $4e$
• C. $2e$
• D. $4e^2$

Answer 1

The correct option is B $4e$

$\because f^{\prime }\left(x\right)=f\left(x\right)\forall x\in R$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=1$
Integrating both sides, w.r.t. $x$, we get
$f\left(x\right)=C{e}^x$
As $f\left(1\right)=2\Rightarrow 2=Ce$
or, $C=\frac{2}{e}$
$\Rightarrow f\left(x\right)=\frac{2}{e}{e}^x$

Now, $h\left(x\right)=f\left(f\right(x\left)\right)$
Differentiating both sides w.r.t. $x$, we get :
$h^{\prime }\left(x\right)=f^{\prime }\left(f\right(x\left)\right)\times f^{\prime }\left(x\right)$
$\therefore h^{\prime }\left(1\right)=f^{\prime }\left(f\right(1\left)\right)\times f^{\prime }\left(1\right)$
$\Rightarrow h^{\prime }\left(1\right)=f^{\prime }\left(2\right)\times f\left(1\right)$
$\Rightarrow h^{\prime }\left(1\right)=\frac{2}{e}{e}^2\times 2$
$\Rightarrow h^{\prime }\left(1\right)=4e$