Problem 29 - Thomas Calculus in SI Units

Question

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 is \( -\operatorname{coth}^{-1} \sqrt{t} - \frac{1}{2\sqrt{t}} \).

Step 1: Identify the Derivative with Respect to the Variable

Given the function, we need to find the derivative of \( y \) with respect to \( t \). This means we will consider \( y = (1-t) \operatorname{coth}^{-1} \sqrt{t} \) and focus on differentiating it with respect to \( t \).

Step 2: Recognize the Product Rule

The function \( y = (1-t) \operatorname{coth}^{-1} \sqrt{t} \) is a product of two functions: \( u = 1-t \) and \( v = \operatorname{coth}^{-1} \sqrt{t} \). We will use the product rule for differentiation: if \( y = uv \), then \( \frac{dy}{dt} = u'v + uv' \).

Step 3: Find the Derivative of the First Function \(u\)

The first function is \( u = 1 - t \). The derivative of \( u \) with respect to \( t \) is \( u' = -1 \).

Step 4: Find the Derivative of the Second Function \(v\)

The second function is \( v = \operatorname{coth}^{-1} \sqrt{t} \). This requires using the chain rule for differentiation. The derivative of \( \operatorname{coth}^{-1} x \) with respect to \( x \) is \( -\frac{1}{1-x^2} \). For \( v = \operatorname{coth}^{-1} \sqrt{t} \), let \( x = \sqrt{t} \), then \( v = \operatorname{coth}^{-1} x \). The chain rule gives \( v' = -\frac{1}{1-x^2} \cdot \frac{d}{dt}(\sqrt{t}) \).

Step 5: Apply the Chain Rule for \(v'\)

Since \( x = \sqrt{t} \), the derivative \( \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}} \). So, \( v' = -\frac{1}{1-(\sqrt{t})^2} \cdot \frac{1}{2\sqrt{t}} = -\frac{1}{1-t} \cdot \frac{1}{2\sqrt{t}} = -\frac{1}{2\sqrt{t}(1-t)} \).

Step 6: Combine Using the Product Rule Formula

Using the derivatives found, substitute into the product rule formula: \( \frac{dy}{dt} = u'v + uv' = (-1)(\operatorname{coth}^{-1} \sqrt{t}) + (1-t) \left(-\frac{1}{2\sqrt{t}(1-t)}\right) \).

Step 7: Simplify the Expression

Simplify the expression to find the final derivative: \( \frac{dy}{dt} = -\operatorname{coth}^{-1} \sqrt{t} - \frac{1}{2\sqrt{t}} \).

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus that's used for differentiating products of two or more functions. Essentially, if you have a function that's the product of two functions, such as \( y = u(t) \cdot v(t) \), the product rule helps us find its derivative. The rule states that:

• If \( y = uv \), then \( \frac{dy}{dt} = u'v + uv' \).

In simpler terms, this means that you take the derivative of the first function and multiply it by the second function, then take the derivative of the second function and multiply it by the first.
For example, with the problem \( y = (1-t) \operatorname{coth}^{-1} \sqrt{t} \), the expression can be separated into two parts:

• \( u = 1-t \)
• \( v = \operatorname{coth}^{-1} \sqrt{t} \)

Now, applying the product rule simplifies the process of differentiation into manageable steps. It's particularly useful for functions where direct application of basic derivative rules is complex.

Chain Rule

The chain rule is another vital tool in calculus, especially when dealing with composite functions. A composite function involves one function inside another, like \( v = \operatorname{coth}^{-1} \sqrt{t} \) in our example.
The chain rule states:

• If a function \( v(t) \) is composed of \( f(g(t)) \), then the derivative \( v'(t) = f'(g(t)) \times g'(t) \).

Think of the chain rule as a way to 'unravel' the layers of a function and differentiate them step by step.
In the case of \( \operatorname{coth}^{-1} \sqrt{t} \), identify \( x = \sqrt{t} \) (inner function) and differentiate \( \frac{d}{dt}(x) = \frac{1}{2\sqrt{t}} \). Combine this with the derivative of the outer function \( \operatorname{coth}^{-1} x \) to find \( v'(t) \).
This breakdown allows us to handle complex expressions by focusing on one component part at a time.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of hyperbolic functions, similar to how inverse trigonometric functions relate to trigonometric functions. In this exercise, the focus is on the inverse hyperbolic cotangent function, \( \operatorname{coth}^{-1}(x) \).
The derivative of \( \operatorname{coth}^{-1}(x) \) with respect to \( x \) is given by:

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

These derivatives are not commonly memorized, but understanding their formulas is essential for applying them correctly in calculus problems.
When involved in a chain rule operation, as seen in this exercise, the derivative of \( \operatorname{coth}^{-1}(x) \) combines with the inner derivative to facilitate the full derivative computation.
This illustrates how powerful calculus tools like the product and chain rules work together with function-specific derivatives to solve complex problems.