$\sum_{k=1}^{\infty} \frac{k}{2^{k}}(x+3)^{k}$

Step 1 ：We are given the series \(\sum_{k=1}^{\infty} \frac{k}{2^{k}}(x+3)^{k}\). This is a power series representation of a function. The general form of a power series is \(\sum_{k=0}^{\infty} a_k (x-c)^k\). In this case, \(a_k = \frac{k}{2^k}\) and \(c = -3\).

Step 2 ：The power series representation of a function can be written as \(f(x) = \sum_{k=0}^{\infty} a_k (x-c)^k\). This is equivalent to \(f(x) = a_0 + a_1(x-c) + a_2(x-c)^2 + a_3(x-c)^3 + \cdots\).

Step 3 ：The derivative of \(f(x)\) is \(f'(x) = a_1 + 2a_2(x-c) + 3a_3(x-c)^2 + 4a_4(x-c)^3 + \cdots\). This can be written as a power series as \(f'(x) = \sum_{k=1}^{\infty} ka_k (x-c)^{k-1}\).

Step 4 ：Comparing this with the given series, we can see that the given series is the derivative of some function \(f(x)\), where \(f(x)\) has a power series representation with \(a_k = \frac{1}{2^k}\) and \(c = -3\).

Step 5 ：So, we need to find the function \(f(x)\) whose derivative is the given series. This can be done by integrating the given series term by term.

Step 6 ：We can observe that the series is a geometric series with common ratio \(\frac{x+3}{2}\). The sum of a geometric series with first term \(a\), common ratio \(r\) and \(n\) terms is given by \(\frac{a(1-r^n)}{1-r}\), and if \(|r| < 1\), the series converges to \(\frac{a}{1-r}\) as \(n\) approaches infinity.

Step 7 ：In this case, \(a = \frac{1}{2}\) and \(r = \frac{x+3}{2}\). So, the sum of the series is \(\frac{\frac{1}{2}}{1-\frac{x+3}{2}} = \frac{1}{2-x}\).

Step 8 ：So, the function \(f(x)\) whose derivative is the given series is \(\int \frac{1}{2-x} dx = -\ln|2-x|\).

Step 9 ：\(\boxed{\text{Final Answer: The function } f(x) \text{ whose derivative is the given series is } f(x) = -\ln|2-x|}\)