If $f$ be a polynomial, then the second derivative of $f({e^x})$is
A. $f'({e^x})$
B. $f''({e^x}){e^x} + f'({e^x})$
C. $f''({e^x}){e^{2x}} + f''({e^x})$
D. $f''({e^x}){e^{2x}} + f'({e^x}){e^x}$

Hint: Use differentiation formula, to find the second derivative of a given function ${e^x}$. As, the second derivative means differentiating the function again and again. Differentiate ${e^x}$with respect to $x$ and again differentiate the first derivative with respect to $x$.

Formula Used:
Differentiation of function –
$\dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}f(x) \times \dfrac{d}{{dx}}(x) = f'(x)$
$\dfrac{d}{{dx}}f'(x) = f''(x)$
Differentiation of exponential function –
$\dfrac{d}{{dx}}({e^x}) = {e^x}$

Complete step by step solution:
Let the given function $f({e^x})$be y
$ \Rightarrow y = f({e^x}) - - - - - (1)$
Differentiate equation (1) with respect to $x$,
$\dfrac{{dy}}{{dx}} = f'({e^x}){e^x} - - - - - (2)$
Differentiate equation (2) with respect to $x$,
$\dfrac{{{d^2}y}}{{d{x^2}}} = {e^x}\left( {\dfrac{d}{{dx}}f'({e^x})} \right) + f'({e^x})\left( {\dfrac{d}{{dx}}({e^x})} \right)$
$\dfrac{{{d^2}y}}{{d{x^2}}} = {e^x}\left( {f''({e^x})} \right){e^x} + f'({e^x}){e^x}$
$\dfrac{{{d^2}y}}{{d{x^2}}} = f''({e^x}){e^{2x}} + f'({e^x}){e^x}$
Hence, the second derivative of $f({e^x})$is $f''({e^x}){e^{2x}} + f'({e^x}){e^x}$.

Option ‘D’ is correct

Note: In such a question first derivate the whole function and then differentiate the function which is inside the main function. When two functions will be in the product apply chain rule (leave the first function outside and differentiate second add with the vice versa). Don’t forget the first case of derivation of the whole function while applying chain rule.