91Ó°ÊÓ

The concept of derivatives is central to calculus, serving as a tool to measure how a function changes as its input changes. To differentiate a function means to find its rate of change or slope at any given point. In this exercise, we focus on applying the chain rule to find the derivatives of specific trigonometric functions.
When given a function like \( \sin(x^2 + 1) \), the chain rule is indispensable. It helps differentiate compositions of functions. Here’s a quick refresher: the chain rule states that to differentiate \( f(g(x)) \), we calculate \( f'(g(x)) \cdot g'(x) \).

• Step 1: Recognize the outer function \( f(u) = \sin(u) \) and the inner function \( u = x^2 + 1 \).
• Step 2: Differentiate the outer function: \( f'(u) = \cos(u) \).
• Step 3: Differentiate the inner function: \( \frac{du}{dx} = 2x \).
• Step 4: Combine results: \( 2x \cos(x^2 + 1) \).

Similarly, for \( \sin(x^3 + 1) \), the chain rule is applied by identifying the inner function \( v = x^3 + 1 \) and its derivative \( 3x^2 \). The final derivative is \( 3x^2 \cos(x^3 + 1) \). These steps illustrate how multiple layers of functions interact under the chain rule.

Antiderivatives are the reverse process of differentiation, often referred to as integration. Finding an antiderivative involves determining the original function given its derivative.
In this context, the focus is on finding the antiderivatives of expressions like \( x \cos(x^2 + 1) \) and \( x^2 \cos(x^3 + 1) \). We leverage the results from the derivatives found earlier to simplify this process.

• Use known derivatives: From our earlier results, we know that \( 2x \cos(x^2 + 1) \) is the derivative of \( \sin(x^2 + 1) \).
• Reverse the derivative: To find the antiderivative of \( x \cos(x^2 + 1) \), we consider the expression derived, determining \( \frac{1}{2} \sin(x^2 + 1) + C \), where \( C \) is an integration constant.
• Similarly, for \( x^2 \cos(x^3 + 1) \), using \( 3x^2 \cos(x^3 + 1) \), we infer the antiderivative as \( \frac{1}{3} \sin(x^3 + 1) + C \).

Understanding these steps enables us to reverse-engineer derivatives back to their original functions, highlighting the interdependent nature of differentiation and integration.

The chain rule is an essential technique in calculus for differentiating composite functions, where one function is nested inside another. It simplifies the process of finding the derivative by breaking it down into manageable parts. Let's delve into why it is useful and how it applies to our example.
Using the chain rule requires identifying the 'outer' and 'inner' functions. In an expression like \( \sin(x^2 + 1) \), \( \sin(u) \) is the outer function, and \( u = x^2 + 1 \) is the inner function.

• Differentiate the outer function with respect to the inner: \( \cos(u) \)
• Differentiate the inner function with respect to \( x \): \( 2x \)
• Multiply these derivatives: \( 2x \cdot \cos(x^2 + 1) \)

This method effectively decomposes the problem, allowing us to handle complex derivatives systematically. Whether the function is \( \sin(x^2 + 1) \) or \( \sin(x^3 + 1) \), the chain rule consistently helps streamline the differentiation process.

Integration by parts is a powerful method in calculus used to integrate products of functions. It can be seen as the reverse of the product rule for differentiation. The formula \[ \int u \, dv = uv - \int v \, du \] provides a systematic way to handle integrals involving products.
In our exercise, integration by parts is used for functions such as \( x \sin(x^2 + 1) \) and \( x^2 \sin(x^3 + 1) \). Here's a step-by-step outline of how it's applied:

• Choose \( u \) and \( dv \): Typically, \( u \) is a polynomial (e.g., \( x \) or \( x^2 \)), and \( dv \) is the remaining part. For \( x \sin(x^2 + 1) \) we set \( u = x \), \( dv = \sin(x^2 + 1) \, dx \).
• Differentiate \( u \): \( du = dx \).
• Integrate \( dv \): Use substitution here to find \( v = -\frac{1}{2}\cos(x^2 + 1) \).
• Apply the integration by parts formula: Resulting in expressions like \( -\frac{1}{2} x \cos(x^2 + 1) + \frac{1}{2} \int \cos(x^2 + 1) \, dx \).

This technique is particularly useful when handling complex integrals that involve trigonometric functions, helping us break down and manage more complicated problems efficiently.