The $n^{t h}$ derivative of $h (x) = e^{3 x + 5} x^{2}$ at $x = 0$ is

e^{5} 3^{n - 2} n (n - 1)

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e^{5} 3^{n + 2} n (n - 1)

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e^{5} 3^{n} n (n - 1)

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e^{5} 3^{n - 2} n (n + 1)

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Solution

The correct option is A $e^{5} 3^{n - 2} n (n - 1)$
Given:
$h (x) = e^{3 x + 5} x^{2}$

Differentiating $h (x)$ both sides, we get
$h^{'} (x) = 3 e^{3 x + 5} x^{2} + 2 x e^{3 x + 5}$
$h^{'} (0) = 0$

Again on differentiating $h^{'} (x)$ , we get
$h^{''} (x) = 3 h^{'} (x) + 6 x e^{3 x + 5} + 2 e^{3 x + 5}$
$h^{''} (x) = 3 h^{'} (x) + 6 x e^{3 x + 5} + 2 e^{3 x + 5}$
$h^{''} (x) = 3 h^{'} (x) + 3 (h^{'} (x) - h (x)) + 2 e^{3 x + 5}$
$h^{''} (x) = 6 h^{'} (x) - 3 h (x) + 2 e^{3 x + 5}$
$h^{''} (0) = 2 e^{5}$
Differenting $h^{''} (x)$ , we get
$h^{'''} (x) = 6 h^{''} (x) - 3 h^{'} (x) + 6 e^{3 x + 5}$
$h^{'''} (0) = 18 e^{5}$
... and so on.

According to the pattern, the value of $n^{t h}$ derivative of $h (x)$ at $x = 0$ is
$h^{n} (0) = e^{5} 3^{n - 2} n (n - 1)$

Hence, option A.

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