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A derivative represents the rate at which a function is changing at any given point. It is the cornerstone of calculus and is used to compute the slope of the function's graph at a specific point.

To find the derivatives of a function, you differentiate it with respect to its variable. For example, the first derivative of a function \( f(x) \) is denoted as \( f'(x) \). This tells us how \( f(x) \) changes when \( x \) changes. The process can be repeated to find higher-order derivatives such as the second, third, and so on.

Let's illustrate derivatives with an example from our problem:

• First, we identified the function \( f(x) = x e^{x^2} \) as the subject of our Taylor series expansion.
• We computed the first derivative using calculus rules (product rule, in this case). For \( f(x) = x e^{x^2} \), the first derivative is \[ f'(x) = e^{x^2}(1 + 2x^2) \].
• Continuing, we calculated higher-order derivatives following similar processes. These steps highlight how the behavior of \( f(x) \) is described by its derivatives at each order.

Understanding derivatives and how to compute them is essential, as they form the basis for constructing a Taylor series.

The product rule is a fundamental technique used in differentiation when dealing with functions that are products of two or more other functions. The rule states that the derivative of a product is not simply the product of the derivatives of the factors.

The product rule formula is:
If \( u(x) = f(x)g(x) \), then \[ u'(x) = f'(x)g(x) + f(x)g'(x). \]

For the given function, \( f(x) = x e^{x^2} \), we apply the product rule:

• First, identify the parts of the product: \( g(x) = x \) and \( h(x) = e^{x^2} \).
• Differentiate each part: The derivative of \( g(x) = x \) is \( g'(x) = 1 \), and the derivative of \( h(x) = e^{x^2} \) is \( h'(x) = 2x e^{x^2} \) (using the chain rule).
• Apply the formula: \[ f'(x) = g'(x)h(x) + g(x)h'(x) = 1 \times e^{x^2} + x \times 2x e^{x^2} = e^{x^2}(1 + 2x^2). \]

Using the product rule efficiently simplifies the process of finding the derivatives of product functions, crucial for tasks such as Taylor series expansion.

Pattern recognition in calculus involves identifying recurring themes and behaviors within mathematical expressions.

In our exercise, pattern recognition helped streamline the Taylor series expansion of the function \( x e^{x^2} \). By investigating the derivatives, we noted:

• The value of \( f(0) \) is 0.
• The first derivative \( f'(0) \) is 1, a non-zero value.
• The second derivative \( f''(0) \) again yields 0.
• Higher-order derivatives followed similar patterns, alternating between non-zero (odd-order) and zero (even-order) at \( x = 0 \).

Recognizing this alternating pattern allowed us to conclude and form the Taylor series without computing every derivative fully. Since even-order derivatives were consistently zero, the series included only terms with odd powers.

Using pattern recognition effectively simplifies complex problems, making it an invaluable tool for finding solutions in a more efficient and intuitive manner.