2. if the function f is defined by f(x)=∑n = 0∞3x2nn!, then f′(x)=? write the first four nonzero terms and…

Explanation:

Step1: Recall power - series derivative rule

If $f(x)=\sum_{n = 0}^{\infty}a_nx^n$, then $f^{\prime}(x)=\sum_{n = 1}^{\infty}na_nx^{n - 1}$. Given $f(x)=\sum_{n=0}^{\infty}\frac{3x^{2n}}{n!}$, we can find its derivative.

Step2: Differentiate the power - series term by term

Let $a_n=\frac{3}{n!}$ and the power of $x$ is $m = 2n$. Then $f^{\prime}(x)=\sum_{n = 1}^{\infty}\frac{3\times2n\times x^{2n-1}}{n!}$. Simplify the coefficient: $\frac{3\times2n}{n!}=\frac{6n}{n!}=\frac{6}{(n - 1)!}$. So $f^{\prime}(x)=\sum_{n = 1}^{\infty}\frac{6x^{2n-1}}{(n - 1)!}$.

Step3: Find the first four non - zero terms

When $n = 1$: $a_1=\frac{6x^{2\times1 - 1}}{(1 - 1)!}=6x$ When $n = 2$: $a_2=\frac{6x^{2\times2 - 1}}{(2 - 1)!}=6x^{3}$ When $n = 3$: $a_3=\frac{6x^{2\times3 - 1}}{(3 - 1)!}=\frac{6x^{5}}{2}=3x^{5}$ When $n = 4$: $a_4=\frac{6x^{2\times4 - 1}}{(4 - 1)!}=\frac{6x^{7}}{6}=x^{7}$

Step4: Find the general term

The general term of the Taylor series of $f^{\prime}(x)$ about $x = 0$ is $a_n=\frac{6x^{2n-1}}{(n - 1)!}$, for $n\geq1$.

Answer:

The first four non - zero terms are $6x,6x^{3},3x^{5},x^{7}$. The general term is $\frac{6x^{2n-1}}{(n - 1)!}$ for $n\geq1$.