Question

If for a continuous function $f,f\left(0\right)=f\left(1\right)=0,f^{\prime }\left(1\right)=2$ and $g\left(x\right)=f\left(e^x\right)e^{f\left(x\right)}$, then $g^{\prime }\left(0\right)$ is equal to

• A. $1$
• B. $2$
• C. $0$
• D. None of these

Answer 1

The correct option is B $2$

Given $g\left(x\right)=f\left(e^x\right)e^{f\left(x\right)}$

Differentiating both sides w.r.t $x$, we get

$g^{\prime }\left(x\right)=f^{\prime }\left(e^x\right).e^x{e}^{f\left(x\right)}+f\left(e^x\right).e^{f\left(x\right)}.f^{\prime }\left(x\right)$ ...(1)

Now, since we already know that $f\left(0\right)=0,f^{\prime }\left(1\right)=2$

Substitute $x=0$ in (1), we get

$g^{\prime }\left(0\right)=f^{\prime }\left(1\right)e^0{e}^{f\left(0\right)}+f\left(e^0\right)e^{f\left(0\right)}{f}^{\prime }\left(0\right)$

$=f^{\prime }\left(1\right)1.e^0+f\left(1\right)e^0{f}^{\prime }\left(0\right)$

$=\mathrm{2.1.1}+0=2$