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Differentiation is like a tool from calculus that helps us understand how things change. Imagine you are tracking how fast a car moves as its speed changes over time. Differentiation is what lets us measure that change. For mathematical functions, it's about finding the rate at which a function is changing at any point. This is done using derivatives.

In the problem at hand, we have a function defined as \( y = \cos x \). To find how the value of \( y \) changes when \( x \) changes, we differentiate the function with respect to \( x \). This process yields the derivative \( \frac{dy}{dx} = -\sin x \).

• The symbol \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \).
• In simpler terms, it tells us how much \( y \) changes for a small change in \( x \).

The Chain Rule is a formula used when we have a composite function—or a function inside another function. It helps us differentiate functions that are connected or nested within each other, like Russian dolls.

In our exercise, we apply the chain rule to find how \( y \) changes over time (\( \frac{dy}{dt} \)). Since \( y \) depends on \( x \), and \( x \) changes over time, the chain rule comes into play.

• According to the chain rule, \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \).
• Here, \( \frac{dy}{dx} \) is the rate at which \( y \) changes with \( x \), and \( \frac{dx}{dt} \) is the rate at which \( x \) changes with time.

Simply multiply the two to get \( \frac{dy}{dt} = -\sin x \times 4 \), where 4 cm/sec is the given rate at which \( x \) changes over time.

Trigonometric functions like sine and cosine are essential in understanding patterns related to angles and rotations. They are especially useful in problems involving cycles and oscillations, such as waves.

In this exercise, the function \( y = \cos x \) involves the cosine function, which is one of the basic trigonometric functions.

• The cosine function is often denoted as \( \cos x \).
• It describes a wave that starts at its maximum value when \( x = 0 \).

For different parts of the exercise, like when \( x \) equals \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \), we substitute these values into the derivative \( -\sin x \times 4 \). This substitution accounts for how specific angles influence the rate of change of the function at those points.

• For \( x = \frac{\pi}{6} \), \( \sin (\frac{\pi}{6}) = \frac{1}{2} \), so \( \frac{dy}{dt} = -2 \text{ cm/sec} \).
• For \( x = \frac{\pi}{4} \), \( \sin (\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), exploding to \( \frac{dy}{dt} = -2\sqrt{2} \text{ cm/sec} \).
• And for \( x = \frac{\pi}{3} \), \( \sin (\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \), turning into \( \frac{dy}{dt} = -2\sqrt{3} \text{ cm/sec} \).

These calculations show how the cosine and sine functions are interconnected, especially when applied to real-world change scenarios.