Question

Let f(x) be a polynomial in x. Then the second derivation of f$\left(e^x\right)$, is:

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

Answer 1

The correct option is D $f"\left(e^x\right)e^{2x}+f^{\prime }\left(e^x\right).e^x$

We have,

$f\left(x\right)$ is a polynomial.

Then,

The second order of $f\left(e^x\right)=?$

Find that,

$\frac{d}{dx}\left[\frac{d}{dx}f\left(e^x\right)\right]=?$

So,

We know that,

$\frac{d}{dx}\left(e^x\right)=e^x$

$So,$

$\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=g^{\prime }\left(x\right)f^{\prime }\left(g\left(x\right)\right)$

Now. Using formula,

$\frac{d}{dx}(I.II)=I.\frac{d}{dx}II+II.\frac{d}{dx}I$

So,

$\frac{d}{dx}\left[f\left(e^x\right)\right]=e^x.f^{\prime }\left(e^x\right)......\left(1\right)$

Again differentiating and we get,

$\frac{d}{dx}\left[\frac{d}{dx}f\left(e^x\right)\right]=\frac{d}{dx}[f^{\prime }\left(e^x\right).e^x]=f^{\prime }\left(e^x\right)e^x+e^x\left[e^x{f}^{\prime \prime }\left(e^x\right)\right]$

$=\frac{d}{dx}\left[\frac{d}{dx}f\left(e^x\right)\right]=e^x{f}^{\prime }\left(e^x\right)+e^{2x}{f}^{\prime \prime }\left(e^x\right)$

Hence, this is the answer.