Use power series to evaluate the function at the given point. Observe in each case that, even with a hand calculator available, it is easier or necessary to use series. \(\frac{d^{5}}{d x^{5}}\left(x^{3} \tan x\right) \quad\) at \(x=0.1\)

To understand power series differentiation effectively, start by grasping the idea of series expansion. This technique allows us to represent complex functions as infinite sums of simpler polynomial terms.
For example, the function \tan(x)\ can be expanded as: \[\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)\]
This series expansion means that instead of dealing with the tangent function directly, we can approximate it using polynomial terms which are easier to work with.
In our exercise, we applied this to \x^3 \tan(x)\ giving us \[x^3 \tan(x) = x^4 + \frac{x^6}{3} + \frac{2x^8}{15} + O(x^{10})\]
Thus, series expansion transforms complex functions into manageable polynomials.

Once we have the series expansion, the next step involves differentiation. Differentiating term-by-term is straightforward because polynomials are simpler to derive. Let's go through this:
First Derivative: \[ \frac{d}{dx}(x^4) = 4x^3 \quad \frac{d}{dx}\bigg(\frac{x^6}{3}\bigg) = 2x^5 \quad \frac{d}{dx}\bigg(\frac{2x^8}{15}\bigg) = \frac{16x^7}{15} \]
Keep going until you have taken the fifth derivative:
Second Derivative: \[ \frac{d^2}{dx^2}(x^4) = 12x^2 \quad \frac{d^2}{dx^2}\bigg(\frac{x^6}{3}\bigg) = 10x^4 \quad \frac{d^2}{dx^2}\bigg(\frac{2x^8}{15}\bigg) = \frac{112x^6}{15} \]
Follow this pattern until you reach the fifth derivative.

Evaluating derivatives at specific points allows you to find the function's rate of change at that point. In our case, we focused on \x = 0.1\. After differentiating five times, we see certain terms involving lower powers become zero: \[ \frac{d^5}{dx^5}\bigg(x^4 \bigg) = 0 \quad \frac{d^5}{dx^5}\bigg(\frac{x^6}{3}\bigg) = 0 \quad \frac{d^5}{dx^5}\bigg(\frac{2x^8}{15}\bigg) = 0 \]
Therefore, terms involving higher powers are critical. When \x = 0.1\, the fifth-order term involving \x^n\ where \ \textless 5\ becomes zero.

The Taylor series is a powerful tool for approximating functions around specific points. The Taylor series expansion of a function \f(x)\ around \x = a\ is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f^{(4)}(a)}{4!}(x-a)^4 + \frac{f^{(5)}(a)}{5!}(x-a)^5 + ... \] In our exercise, expanding the function \x^3 \tan(x)\ around \x=0\ simplifies our calculation for higher-order derivatives. Taylor series help break down functions into simpler polynomial components, making differentiation and evaluation more manageable.