Question

Use the Taylor’s series to obtain 𝑛 𝑡ℎ derivative at 𝑥 = 0 (i.e. 𝑑 𝑛𝑦 𝑑𝑛𝑥 at 𝑥 = 0) of 𝑦 = 𝑥 3𝑒 �

Answer 1

The $n$-th derivative of $y = x^3e^x$ at $x=0$ is: $y^{(n)}(0) = \begin{cases} 0 & \text{if } n < 3 \\ n(n-1)(n-2) & \text{if } n \ge 3 \end{cases}$

Answer 2

The Taylor series for $e^x$ is multiplied by $x^3$ to obtain the series for $y=x^3e^x$. By re-indexing the resulting series to match the form $\sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}x^n$, we equate the coefficients of $x^n$. This directly gives the value of $y^{(n)}(0)$ for different values of $n$. For $n<3$, the coefficients are zero, and for $n \ge 3$, the coefficients yield $n(n-1)(n-2)$.