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The second derivative of a function provides us with valuable insights into the concavity and acceleration aspects of that function. When solving differential equations, knowing the second derivative allows us to address equations involving the function and its derivatives, like the one in our exercise: \(y'' + y = \sin x\).

A second derivative, denoted as \(y''\), is simply the derivative of the first derivative, \(y'\). This means we have to differentiate the function twice. For instance, if \(y = \sin x\), the first derivative is \(y' = \cos x\), and differentiating \(\cos x\) gives us \(y'' = -\sin x\). This recursive process is applied to find the second derivatives in steps 1 through 4 of the solution.

When working with differential equations like \(y'' + y = \sin x\), determining the second derivative helps verify if substituting \(y''\) and \(y\) actually equals \(\sin x\). If they do, the function in question is a solution to the equation.

The Product Rule is a technique used in calculus to find the derivative of a product of two functions. If you have functions \(u(x)\) and \(v(x)\), the rule states that their derivative, \( (uv)' \), is given by \( u'v + uv' \), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively.

In our exercise, Options (c) and (d) require the application of the Product Rule to compute the derivatives. For example, with \(y = \frac{1}{2}x \sin x\), the function is a product of \(u = \frac{1}{2}x\) and \(v = \sin x\).

• First, find the derivative of \(u\): \(u' = \frac{1}{2}\).
• Next, find the derivative of \(v\): \(v' = \cos x\).
• Apply the Product Rule: \(y' = u'v + uv' = \frac{1}{2}\sin x + \frac{1}{2}x\cos x\).

This method ensures we correctly differentiate products of functions and is crucial in finding whether such a function satisfies a differential equation.

Trigonometric functions such as \(\sin x\) and \(\cos x\) occur frequently in mathematics, particularly when dealing with periodic phenomena. These functions are foundational in calculus and crucial when solving differential equations.

In the context of our differential equation \(y'' + y = \sin x\), trigonometric functions play a dual role. They are part of the potential solutions, as seen in the exercise options, and appear in the equation itself. This provides a natural function type that is likely to be part of the solution.

When differentiating trigonometric functions, it is essential to remember the following basic derivatives:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).

Understanding these derivatives is key, as several steps rely on accurate differentiation of trigonometric functions. This foundation helps determine if the function is a valid solution by calculating the second derivative and substituting back into the original differential equation.