Problem 30 - Thomas Calculus

Question

In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\left(1-t^{2}\right) \cot h^{-1} t$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = -1 - 2t \coth^{-1}(t) \).

Step 1: Recall the Chain Rule

To find the derivative of a composite function, we use the chain rule. If we have a function of the form \( y = u imes v \), the derivative \( \frac{dy}{dt} \) can be found using the product rule as \( \frac{dy}{dt} = u'v + uv' \). In this problem, \( u = 1-t^2\) and \( v = \coth^{-1}(t) \).

Step 2: Differentiate \( u = 1-t^2 \)

We first need to differentiate the function \( u = 1-t^2 \) with respect to \( t \). The derivative is \( u' = -2t \).

Step 3: Find the Derivative of \( v = \coth^{-1}(t) \)

The inverse hyperbolic cotangent function \( \coth^{-1}(t) \) has the derivative \( v' = -\frac{1}{1-t^2} \) for \( |t| > 1 \).

Step 4: Apply the Product Rule

Subsequently, using the derivatives found, apply the product rule: \( \frac{dy}{dt} = (1-t^2)(-\frac{1}{1-t^2}) + (-2t)\coth^{-1}(t) \).

Step 5: Simplify the Expression

Simplifying the first term of the product rule gives \( -1 \). Therefore, the derivative simplifies to \( \frac{dy}{dt} = -1 - 2t \coth^{-1}(t) \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. A composite function is essentially a function inside another function, like nesting dolls. If we have a function composed of two functions, such as \( y = f(g(t)) \), then the chain rule helps us determine the derivative by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function. This could be expressed as \( f'(g(t)) \times g'(t) \).
The idea is to "chain" the rate of change of the outer function with the rate of change of the inner function. In many real-world problems, we encounter situations like this, where one event or action influences another. The chain rule makes solving these problems manageable by breaking them down into simpler derivatives.

Product Rule

The product rule is another important differentiation rule that deals with the derivative of the product of two functions. If we have two functions \( u(t) \) and \( v(t) \) multiplying each other, the derivative of their product \( y = u(t) \cdot v(t) \) is found by:
\[\frac{dy}{dt} = u'(t)v(t) + u(t)v'(t)\]
Here, we essentially keep one function constant while differentiating the other, and then we switch roles. This ensures we capture all possible ways the two functions could be changing as they multiply together. The product rule is incredibly handy in situations where two quantities are interacting, altering how they both behave over time.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as \( \coth^{-1}(t) \), are the "inverse" of hyperbolic functions like \( \coth(t) \). These functions are used in many fields, including engineering and physics, where equations need hyperbolic instead of circular characteristics. The derivatives of these functions have specific formulas, often involving expressions like \( 1 - t^2 \) underneath the square root in the denominators, representing hyperbolic identities. For instance, the derivative of \( \coth^{-1}(t) \) is \(-\frac{1}{1-t^2}\) for \(|t| > 1\).
These derivatives are essential when working with hyperbolic equations or when dealing with complex transformations where hyperbolic functions are natural fits.

Differentiation

Differentiation is a calculus process for finding a function's derivative, which describes the rate at which inputs change outputs. It answers the question, "How does a small change in my input affect my output?" The result of differentiation offers insight into the behavior of a function in terms of slope, rates, and tangents.
In practice, differentiation rules like the chain rule and product rule are often used together, as functions can be complex amalgamations of multiple nested or multiplied functions. Consider these rules as your mathematical toolkit, each with a specified use, allowing you to effectively navigate and solve derivatives across different types of functions.

Composite Functions

Composite functions are formed by substituting one function inside another. These functions are denoted as \( f(g(x)) \), where \( g(x) \) is nested within \( f(x) \). The challenge and beauty of composite functions lie in their layers.
When differentiating composite functions, we typically use the chain rule. These functions are ubiquitous in mathematics, representing a sequence of functions acting on the original input to give the final output. They highlight real-world scenarios where multiple processes or stages transform an initial input, making understanding them essential for any comprehensive study of calculus.