Question

If $f\left(x\right)$ is a polynomial in $x$, then the second derivative of $f\left(e^x\right)$ at $x=1$ is

• A. $e{f}^{\prime \prime }\left(e\right)+f^{\prime }\left(e\right)$
• B. $\left(f^{\prime \prime }\right(e)+f^{\prime }(e\left)\right)e^2$
• C. $e^2{f}^{\prime \prime }\left(e\right)$
• D. $\left(f^{\prime \prime }\right(e)e+f^{\prime }(e\left)\right)e$

Answer 1

The correct option is D $\left(f^{\prime \prime }\right(e)e+f^{\prime }(e\left)\right)e$

$\frac{d}{dx}\left(f\right(e^x\left)\right)=f^{\prime }\left(e^x\right)\times \frac{d}{dx}\left(e^x\right)=f^{\prime }\left(e^x\right)\cdot e^x$
Similarly,
$\frac{d^2}{d{x}^2}\left(f\right(e^x\left)\right)=f^{\prime \prime }\left(e^x\right)\frac{d}{dx}\left(e^x\right)e^x+\frac{d}{dx}\left(e^x\right)f^{\prime }\left(e^x\right)=f^{\prime \prime }\left(e^x\right)\cdot e^{2x}+f^{\prime }\left(e^x\right)\cdot e^x$
At $x=1$,
$\frac{d^2}{d{x}^2}\left(f\right(e^x\left)\right)=\left(f^{\prime \prime }\right(e)e+f^{\prime }(e\left)\right)e$