Explanation
We have,
f(x) is a polynomial.
Then,
The second order of f(ex)=?
Find that,
ddx[ddxf(ex)]=?
So,
We know that,
ddx(ex)=ex
ddx[f(g(x))]=g′(x)f′(g(x))
Now. Using formula,
ddx(I.II)=I.ddxII+II.ddxI
ddx[f(ex)]=ex.f′(ex)......(1)
Again differentiating and we get,
ddx[ddxf(ex)]=ddx[f′(ex).ex]=f′(ex)ex+ex[exf″
=\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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