Explanation
We have,
f\left( x \right) is a polynomial.
Then,
The second order of f\left( {{e}^{x}} \right)=?
Find that,
\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right]=?
So,
We know that,
\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}
\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=g'\left( x \right)f'\left( g\left( x \right) \right)
Now. Using formula,
\dfrac{d}{dx}\left( I.II \right)=I.\dfrac{d}{dx}II+II.\dfrac{d}{dx}I
\dfrac{d}{dx}\left[ f\left( {{e}^{x}} \right) \right]={{e}^{x}}.f'\left( {{e}^{x}} \right)\,\,\,......\,\,\left( 1 \right)
Again differentiating and we get,
\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right]=\dfrac{d}{dx}\left[ f'\left( {{e}^{x}} \right).{{e}^{x}} \right]=f'\left( {{e}^{x}} \right){{e}^{x}}+{{e}^{x}}\left[ {{e}^{x}}f''\left( {{e}^{x}} \right) \right]
=\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right]={{e}^{x}}f'\left( {{e}^{x}} \right)+{{e}^{2x}}f''\left( {{e}^{x}} \right)
Hence, this is the answer.
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