Problem 29 - Thomas Calculus

Question

In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=(1-t) \operatorname{coth}^{-1} \sqrt{t} $$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Structure of the Function

The function given is\[ y = (1-t) \coth^{-1}(\sqrt{t}) \]where \( \coth^{-1} \) is the inverse hyperbolic cotangent, and \( \sqrt{t} \) is the composition inside the inverse cotangent. We need to find the derivative of this function with respect to \( t \). This requires applying product rule and chain rule.

Step 2: Apply the Product Rule

The product rule states that if \( y = u \cdot v \), then \( \frac{dy}{dt} = u'v + uv' \). For our function, let\[ u = 1-t \quad \text{and} \quad v = \coth^{-1}(\sqrt{t}) \]Apply the product rule to find \( \frac{dy}{dt} \) by calculating \( u' \), \( v' \), and their products.

Step 3: Differentiate \( u = 1 - t \)

The derivative of \( u = 1 - t \) with respect to \( t \) is straightforward:\[ u' = -1 \]

Step 4: Differentiate \( v = \coth^{-1}(\sqrt{t}) \) Using Chain Rule

To differentiate \( v = \coth^{-1}(\sqrt{t}) \), use the chain rule:Firstly, recall that the derivative of \( \coth^{-1}(x) \) with respect to \( x \) is\[ \frac{d}{dx} \coth^{-1}(x) = -\frac{1}{1 - x^2} \]Since \( x = \sqrt{t} \), set \( w = \sqrt{t} \), then\[ v = \coth^{-1}(w) \quad \text{where} \quad w = \sqrt{t} \]Differentiate \( w = \sqrt{t} \) to get \( \frac{dw}{dt} = \frac{1}{2\sqrt{t}} \), and apply the chain rule:\[ v' = -\frac{1}{1-(\sqrt{t})^2} \cdot \frac{1}{2\sqrt{t}}\]

Step 5: Simplify the Derivative \( v' \)

Simplify the derivative \( v' \) obtained in Step 4:Given:\[ 1-(\sqrt{t})^2 = 1-t \]Then,\[ v' = -\frac{1}{1-t} \cdot \frac{1}{2\sqrt{t}} = -\frac{1}{2\sqrt{t}(1-t)} \]

Step 6: Compute the Final Derivative \( \frac{dy}{dt} \)

Plug \( u' = -1 \), \( v = \coth^{-1}(\sqrt{t}) \), \( u = 1-t \), and \( v' = -\frac{1}{2\sqrt{t}(1-t)} \) into the product rule:\[\frac{dy}{dt} = (-1) \cdot \coth^{-1}(\sqrt{t}) + (1-t) \cdot \left( -\frac{1}{2\sqrt{t}(1-t)} \right)\]Simplify this by combining terms:\[\frac{dy}{dt} = - \coth^{-1}(\sqrt{t}) - \frac{1}{2\sqrt{t}}\]

Key Concepts

Understanding the Product Rule for Derivatives

In calculus, when dealing with the differentiation of products of functions, the **product rule** is indispensable. The product rule is applied when we have a function that is the product of two other functions. Let's break it down simply: If you have two functions, say \( u(t) \) and \( v(t) \), multiplied together as \( y = u(t) \cdot v(t) \), the derivative of \( y \) with respect to \( t \) is given by:

• \( \frac{dy}{dt} = u'(t)v(t) + u(t)v'(t) \)

We need to compute the derivatives of both functions separately and then plug them back using the product rule formula. In our scenario, \( u(t) = 1-t \) and \( v(t) = \coth^{-1}(\sqrt{t}) \).
By finding both derivatives \( u'(t) \) and \( v'(t) \), we can proceed to apply the product rule. Knowing this will open various doors to solving many real-life problems involving rates of change in products of quantities.

Mastering the Chain Rule in Calculus

The **chain rule** is a powerful technique in calculus used to differentiate composite functions. A composite function is one where a function is composed inside another function, like \( f(g(t)) \). The chain rule states:

• \( \frac{d}{dt} f(g(t)) = f'(g(t)) \cdot g'(t) \)

Here, you first take the derivative of the outer function \( f \) with respect to its argument \( g(t) \), and then multiply it by the derivative of the inner function \( g \) with respect to \( t \).
Using the chain rule, we tackled the differentiation of \( v(t) = \coth^{-1}(\sqrt{t}) \). Since \( \sqrt{t} \) is inside the \( \coth^{-1} \) function, it becomes our inner function.
We first find the derivative of \( \coth^{-1}(x) \) and then multiply it by the derivative of \( \sqrt{t} \) to get the complete derivative, effectively linking the chain of differentiation processes. This systematic approach is crucial in handling layers of functions in complex derivatives.

Exploring Inverse Hyperbolic Functions

Inverse hyperbolic functions, like \( \coth^{-1}(x) \), are related to hyperbolic functions which can be seen as analogs of trigonometric functions but for hyperbolas instead of circles. They are especially useful in calculus and modeling real-world phenomena.
The function \( \coth^{-1}(x) \), or inverse hyperbolic cotangent, is the function whose hyperbolic cotangent is \( x \). The derivative of \( \coth^{-1}(x) \) with respect to \( x \) is unique and important:

• \( \frac{d}{dx} \coth^{-1}(x) = -\frac{1}{1 - x^2} \)

This formula illustrates how inverse hyperbolic functions behave differently compared to standard trigonometric functions in terms of their derivatives.
In our problem, the inverse hyperbolic function adds complexity and requires careful application of both product and chain rules to accurately find the resulting composite derivative. Getting comfortable with these functions extends your fluency in applying calculus to a broad range of problems.