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The chain rule is a fundamental technique in differentiation used when dealing with composite functions, that is, functions within functions. It essentially states that if you have a function \( y = f(g(x)) \), the derivative \( y' \) is computed as the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function:\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]In the context of our exercise, we apply it when dealing with the natural logarithm \( \ln(t^2 + 4) \). Here, \( t^2 + 4 \) is the inner function, \( u \), and \( \ln(u) \) is the outer function. The derivative of the natural logarithm function \( \ln(u) \) is \( \frac{u'}{u} \).

Applying the chain rule helps to break down complex expressions into manageable parts, allowing us to differentiate each component effectively. For instance, identifying \( u = t^2 + 4 \) allows us to focus on finding \( u' = 2t \), simplifying the differentiation of \( \ln(t^2 + 4) \) to \( \frac{2t}{t^2 + 4} \). Thus, understanding and applying the chain rule is crucial to finding derivatives of functions composed of multiple layers.

The derivative of logarithmic functions plays a pivotal role in calculus, especially when the function involves natural logarithms \( \ln(x) \). The general rule is that the derivative of \( \ln(u) \), with respect to \( x \), is \( \frac{u'}{u} \).
This formula is particularly useful because it provides a straightforward way to differentiate functions that have logarithms. We often see this in equations involving exponential growth or decay, engineering, and natural processes.

For the given problem, \( y_1 = \ln(t^2 + 4) \), we identify the inner function \( u = t^2 + 4 \), and find its derivative \( u' = 2t \). Using the derivative of logarithmic functions rule, the derivative of \( \ln(t^2 + 4) \) becomes \( \frac{2t}{t^2 + 4} \). This transformation highlights how the logarithmic differentiation splits the problem into finding simpler derivatives of basic functions.

Inverse trigonometric functions like \( \arctan(x) \) have unique derivatives that require careful handling, especially when they're part of more complex expressions. The general derivative of \( \arctan(u) \), where \( u \) is a function of \( x \), is \( \frac{u'}{1 + u^2} \).
In our example, we're differentiating \( y_2 = -\frac{1}{2} \arctan(t/2) \). Identifying \( u = t/2 \), we first compute its derivative, \( u' = 1/2 \). Applying the derivative rule for \( \arctan(x) \), the derivative of \( -\frac{1}{2} \arctan(t/2) \) turns into \(-\frac{1}{2} \times \frac{1/2}{1 + (t/2)^2} = -\frac{1}{4(1 + t^2/4)}\).

These derivatives are instrumental when solving problems involving inverse trigonometric functions, as they enable understanding of changes in angles and their impacts in related fields such as physics and engineering. Thus, knowing how to compute them allows you to tackle a variety of calculus problems effectively.