consider the function f(x)=5 tan^(-1)(x^2). 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 $\tan^{-1}(t)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}t^{2n+1}=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\frac{t^{7}}{7}+\cdots$, for $|t|\lt1$.

Step2: Substitute $t=x^{2}$ into the Taylor - series of $\tan^{-1}(t)$

We have $f(x)=5\tan^{-1}(x^{2})=5\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}(x^{2})^{2n+1}=5\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{4n + 2}=5x^{2}-\frac{5x^{6}}{3}+\frac{5x^{10}}{5}-\frac{5x^{14}}{7}+\cdots$.

Step3: Differentiate the Taylor - series of $f(x)$ term - by - term

Using the power - rule $\frac{d}{dx}(x^{m})=mx^{m - 1}$, we get $f^\prime(x)=\frac{d}{dx}\left(5\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{4n+2}\right)=5\sum_{n = 0}^{\infty}\frac{(-1)^{n}(4n + 2)}{2n + 1}x^{4n+1}$. Simplifying the coefficient $\frac{(-1)^{n}(4n + 2)}{2n + 1}=2(-1)^{n}(2n + 1)\div(2n + 1)=2(-1)^{n}$, so $f^\prime(x)=10\sum_{n = 0}^{\infty}(-1)^{n}x^{4n + 1}=10x-10x^{5}+10x^{9}-10x^{13}+\cdots$.

Answer:

C. $10x - 10x^{5}+10x^{9}-10x^{13}+\cdots$