Question

The $n^{th}$ derivative of $x{e}^x$ vanishes when

• A. $x=0$
• B. $x=-1$
• C. $x=-n$
• D. $x=n$

Answer 1

The correct option is C

$x=-n$

Find the value $n^{th}$ derivative doesn't exists

Let $y=x{e}^x$

$$\begin{gathered} \frac{dy}{dx}=x{e}^x+e^x \\ \frac{d^{2}y}{d{x}^2}=x{e}^x+e^x+e^x=x{e}^x+2e^x \\ \frac{d^{3}y}{d{x}^3}=x{e}^x+3e^x \end{gathered}$$

Similarly $n^{th}$ derivative is $x{e}^x+n{e}^x$

$n^{th}$ derivative vanishes

$$\begin{gathered} \Rightarrow x{e}^x+n{e}^x=0 \\ \Rightarrow e^x\left(x+n\right)=0 \\ \therefore x=-n \end{gathered}$$

Hence, the correct option is $\left(\mathrm{C}\right)$.