Obtain the Maclaurin series expansion for \(f(x)=\cosh x\).

First, identify the function we are working with: \(f(x) = \cosh x\). The hyperbolic cosine function has the following property: $$\cosh x = \frac{e^x + e^{-x}}{2}$$ Our first step will be to differentiate 𝑓(𝑥) multiple times to find the general derivative.

Let's represent the n-th derivative of f(x) by \(f^{(n)}(x)\). Consider the following derivatives: 1. \(f'(x) = \frac{d}{dx}(\cosh x) = \frac{d}{dx}(\frac{e^x + e^{-x}}{2}) = \sinh x\) 2. \(f''(x) = \frac{d^2}{dx^2}(\cosh x) = \frac{d^2}{dx^2}(\frac{e^x + e^{-x}}{2}) = \cosh x\) 3. \(f^{(3)}(x) = \frac{d^3}{dx^3}(\cosh x) = \frac{d^3}{dx^3}(\frac{e^x + e^{-x}}{2}) = \sinh x\) 4. \(f^{(4)}(x) = \frac{d^4}{dx^4}(\cosh x) = \frac{d^4}{dx^4}(\frac{e^x + e^{-x}}{2}) = \cosh x\) Notice that after differentiating the function four times, we are back to the original function (i.e., 𝑓(𝑥) = cosh 𝑥). From this point, the derivatives will follow the same pattern. Therefore, the general form of the n-th derivative can be written as: $$f^{(n)}(x) = \begin{cases} \cosh x & \text{if n is even} \\ \sinh x & \text{if n is odd} \end{cases}$$

Now that we have the general form of the derivatives, we need to evaluate them at x=0. If n is even: $$f^{(n)}(0) = \cosh 0 = 1$$ If n is odd: $$f^{(n)}(0) = \sinh 0 = 0$$

Using the values for the n-th derivatives of 𝑓(𝑥) at x=0, we can now construct the Maclaurin series expansion using the general formula: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$ Since the derivatives evaluated at x=0 are either 1 or 0, the sum can be simplified: $$f(x) = \cosh x = \sum_{n=0}^{\infty} \frac{1}{(2n)!} x^{2n}$$ Here, we used the fact that the odd derivative terms will be zero, so we only include even terms in the sum. The final answer for the Maclaurin series expansion of \(f(x)=\cosh x\) is: $$f(x) = \cosh x = \sum_{n=0}^{\infty} \frac{1}{(2n)!} x^{2n}$$