91Ó°ÊÓ

The chain rule is a fundamental concept in calculus that helps us differentiate composite functions. Suppose you have a function defined as a composition of two functions, say \( y = f(g(x)) \). The chain rule states that to find the derivative of this composition, you must take the derivative of \( f \) with respect to \( g(x) \) and multiply it by the derivative of \( g(x) \) with respect to \( x \).Essentially, the chain rule can be written as:

• \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \)

In our exercise, though the function is not composite, we treat the differentiation with respect to time, \( t \), by using the chain rule. We differentiate the function \( y = \tan(x) \) to get \( dy/dx = \sec^2(x) \), and then apply the chain rule:

• \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

Here, \( \frac{dx}{dt} = 3 \) feet per second, which represents the rate at which \( x \) changes with respect to time, \( t \). Using this approach allows us to solve for \( \frac{dy}{dt} \), or how \( y \) changes over time.

Trigonometric functions are functions that relate the angles of a triangle to the lengths of its sides. Among these functions, tangent (tan) and secant (sec) are particularly important here. When differentiating trigonometric functions, each has its own unique derivative:

• The derivative of \( \tan(x) \) is \( \sec^2(x) \).
• \( \sec(x) \) is a reciprocal trigonometric function which is \( 1/\cos(x) \).

In the context of our problem, we're dealing with \( y = \tan(x) \). This choice leads to the derivative \( dy/dx = \sec^2(x) \) due to the specific differentiation rules of trigonometric functions. This derivative is crucial for finding how the output of the function changes as \( x \) changes, especially when applying the chain rule.

The rate of change is an important concept in calculus, especially in real-world applications involving motion and growth. It describes how a quantity changes over time. For the problem at hand, we're particularly interested in how both \( x \) and \( y = \tan(x) \) change over time, \( t \).In this exercise:

• \( \frac{dx}{dt} = 3 \) feet per second is given, representing how quickly \( x \) is changing with respect to time.
• \( \frac{dy}{dt} \) is what we're trying to find — the rate at which \( y \), or \( \tan(x) \), changes as \( x \) changes with respect to time.

To calculate \( \frac{dy}{dt} \), we use the chain rule, which combines these rates of change:

• \( \frac{dy}{dt} = \sec^2(x) \cdot \frac{dx}{dt} \)

By substituting different values of \( x \) — \(-\pi/3\), \(-\pi/4\), and \(0\) — we find different rates of change for \( y \) at these points. These results show how the speed of tan function's change can vary greatly based on the input \( x \).