We are given two functions $f(x)$, and we need to find the power series expansions for the derivative $f'(x)$ and $g(x) = \int_0^x f(u) \, du$, where $f(u)$ is the original function in each case. Let's approach both problems step by step.

Part 1: $f(x) = (1 + x)^{-1}$

The given power series expansion for $f(x)$ is: $f(x) = (1 + x)^{-1} = \sum_{n=0}^{\infty} (-1)^n x^n$

Step 1: Find $f'(x)$

We can differentiate term by term:

= \sum_{n=0}^{\infty} (-1)^n \cdot n x^{n-1} = \sum_{n=1}^{\infty} (-1)^n \cdot n x^{n-1}$$ Thus, the power series expansion for $$f'(x)$$ is: $$f'(x) = \sum_{n=1}^{\infty} (-1)^n \cdot n x^{n-1}$$ #### Step 2: Find $$g(x) = \int_0^x f(u) \, du$$ Now, integrate the power series of $$f(x)$$ term by term: $$g(x) = \int_0^x \left( \sum_{n=0}^{\infty} (-1)^n u^n \right) du = \sum_{n=0}^{\infty} (-1)^n \int_0^x u^n \, du = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$$ So the power series expansion for $$g(x)$$ is: $$g(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$$ ### Part 2: $$f(x) = \frac{\sin(x)}{x}$$ The given power series expansion for $$f(x)$$ is: $$f(x) = \frac{\sin(x)}{x} = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!}$$ #### Step 1: Find $$f'(x)$$ Differentiate term by term: $$f'(x) = \frac{d}{dx} \left( \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!} \right) = \sum_{k=0}^{\infty} (-1)^k \frac{2k \cdot x^{2k-1}}{(2k+1)!}$$ Thus, the power series expansion for $$f'(x)$$ is: $$f'(x) = \sum_{k=1}^{\infty} (-1)^k \frac{2k \cdot x^{2k-1}}{(2k+1)!}$$ #### Step 2: Find $$g(x) = \int_0^x f(u) \, du$$ Now, integrate term by term: $$g(x) = \int_0^x \left( \sum_{k=0}^{\infty} (-1)^k \frac{u^{2k}}{(2k+1)!} \right) du = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(2k+1)!} \int_0^x u^{2k} \, du = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1) \cdot (2k+1)!}$$ So the power series expansion for $$g(x)$$ is: $$g(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1) \cdot (2k+1)!}$$ ### Summary: For $$f(x) = (1 + x)^{-1}$$: - $$f'(x) = \sum_{n=1}^{\infty} (-1)^n n x^{n-1}$$ - $$g(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$$ For $$f(x) = \frac{\sin(x)}{x}$$: - $$f'(x) = \sum_{k=1}^{\infty} (-1)^k \frac{2k \cdot x^{2k-1}}{(2k+1)!}$$ - $$g(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1) \cdot (2k+1)!}$$ Would you like further details or clarifications on any part of this process? Here are 5 related questions to extend your understanding: 1. How would you find the radius of convergence for the series expansion of $$f'(x)$$? 2. Can you find a closed-form expression for $$g(x)$$ in the case of $$f(x) = (1 + x)^{-1}$$? 3. How does the series expansion of $$\frac{\sin(x)}{x}$$ differ from that of $$\sin(x)$$? 4. What happens to the series expansion of $$g(x)$$ if $$f(x) = \cos(x)$$ instead of $$\sin(x)/x$$? 5. How would the series expansions change if $$f(x) = (1 - x)^{-1}$$? **Tip:** When differentiating or integrating power series, ensure that the operations are valid within the interval of convergence.