Problem 6 Find the rate of change of \(y\)... [FREE SOLUTION] (2024)

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Understanding the chain rule is essential for solving complex calculus problems that involve compounded functions. In layman's terms, the chain rule helps us find the derivative of a function within a function—like opening a set of nested dolls. When the rate of change of a function, let's call it 'g', affects another function 'f', the rate of change of the composite function 'f(g(t))' can be found by multiplying the derivative of the outside function 'f' with respect to 'g', by the derivative of the inside function 'g' with respect to 't'.

For example, if we have a function like in exercise (a), where y equals the sine of the reciprocal of t, or in mathematical terms, y = sin(1/t), applying the chain rule helps us navigate the intertwined relationship between the sine function and 1/t. It is clear as day that by understanding the chain rule, students can adeptly tackle the derivatives of composite functions, making it a potent tool in the calculus toolkit.

When division comes into play in calculus, the quotient rule is the superhero that saves the day. It allows us to differentiate functions that are fractions of each other, where one function is divided by another. The formula can be expressed as: if y = u/v, then the derivative of y with respect to t is dy/dt = (v * u' - u * v') / v^2, where u and v are functions of t, and u' and v' are their respective derivatives.

Let’s consider the scenario in exercise (d), where y is a ratio of two polynomials: (1+t+t2) / (1-t). To find the derivative, we identify u and v, calculate their derivatives, and plugin them into the quotient formula. This systematic approach can be applied to any ratio of functions, simplifying an otherwise daunting calculus problem. Remembering and applying the quotient rule is invaluable when deciphering the change of rates in rational expressions.

Exponential functions, wherein a constant base is raised to a variable exponent, often emerge in growth and decay models, among other applications. Differentiating these functions requires us to remember that the derivative of an exponential function with base e (Euler's number), e^t, is wonderfully the same as the function itself, e^t. This property is what makes e so remarkable in calculus.

In exercise (e), we deal with a more intricate situation that involves both an exponential function and a quotient. Here, the derivative of e^t remains unaffected as e^t, but the next steps involve the quotient rule, as the exponential function is divided by another function. This illustrates the layered nature of calculus problems where multiple rules might apply, showing the power of understanding the derivative of exponential functions as a foundational skill.

Hyperbolic functions may sound like a high-level mathematical concept, but they can be understood as close relatives of the familiar trigonometric sine and cosine functions, albeit with a twist. Specifically, the function sinh(t), known as the hyperbolic sine of t, aligns itself with exponential growth and decay scenarios. Its derivative, cosh(t) or the hyperbolic cosine of t, remains ever-present in calculating rates of change involving sinh(t).

For instance, in exercise (c), y equals sinh(t2), which is a composition of a hyperbolic function and a power function. Utilizing the chain rule, we differentiate t2 as 2t and then multiply by the derivative of the hyperbolic sine, which is the hyperbolic cosine. Awareness of the derivatives of hyperbolic functions empowers students to dive into problems involving rates of change with these exotic, but increasingly applicable, mathematical functions.