consider the function f(x)=tan^(-1)(x^5). a. differentiate the taylor series centered at 0 for f(x). b…

Explanation:

Step1: Recall the Taylor - series of $\tan^{-1}(t)$

The Taylor - series of $\tan^{-1}(t)$ centered at $t = 0$ is given by $\tan^{-1}(t)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}t^{2n+1}=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\frac{t^{7}}{7}+\cdots$, for $|t|\lt1$. Let $t=x^{5}$, then $f(x)=\tan^{-1}(x^{5})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}(x^{5})^{2n+1}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{10n + 5}=x^{5}-\frac{x^{15}}{3}+\frac{x^{25}}{5}-\frac{x^{35}}{7}+\cdots$.

Step2: Differentiate the Taylor - series of $f(x)$

Using the power - rule for differentiation $\frac{d}{dx}(x^{m})=mx^{m - 1}$, we differentiate the series term - by - term. If $y=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{10n+5}$, then $y^\prime=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(10n + 5)}{2n + 1}x^{10n+4}$. Simplifying $\frac{10n + 5}{2n + 1}=5$, so $y^\prime=\sum_{n = 0}^{\infty}(-1)^{n}5x^{10n + 4}=5x^{4}-5x^{14}+5x^{24}-5x^{34}+\cdots$.

Answer:

C. $5x^{4}-5x^{14}+5x^{24}-5x^{34}+\cdots$