Question

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

• A. $e^5{3}^{n-2}n(n-1)$
• B. $e^5{3}^{n+2}n(n-1)$
• C. $e^5{3}^{n}n(n-1)$
• D. $e^5{3}^{n-2}n(n+1)$

Answer 1

The correct option is A $e^5{3}^{n-2}n(n-1)$

Given:

$h\left(x\right)=e^{3x+5}{x}^2$

Differentiating $h\left(x\right)$ both sides, we get

$h^{\prime }\left(x\right)=3e^{3x+5}{x}^2+2x{e}^{3x+5}$
$h^{\prime }\left(0\right)=0$

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

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

Hence, option A.