Problem 30 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is \( y' = -2t \coth^{-1}(t) - 1 \).

Step 1: Identify the Function Components

We have a function defined as \( y = (1-t^2) \coth^{-1}(t) \). The function can be seen as a product of two functions: \( u(t) = 1-t^2 \) and \( v(t) = \coth^{-1}(t) \). This suggests that we will likely use the product rule for differentiation.

Step 2: Apply the Product Rule for Differentiation

Using the product rule, the derivative of \( y = u(t)v(t) \) is \( y' = u'v + uv' \). We need to find both \( u'(t) \) and \( v'(t) \) to proceed.

Step 3: Differentiate Each Function Component

Differentiate \( u(t) = 1 - t^2 \). \[ u'(t) = \frac{d}{dt}(1-t^2) = -2t \]Next, differentiate \( v(t) = \coth^{-1}(t) \).\[ v'(t) = \frac{d}{dt}(\coth^{-1}(t)) = \frac{-1}{1-t^2} \] (using the standard derivative for the inverse hyperbolic cotangent function).

Step 4: Substitute Derivatives into the Product Rule Formula

Substitute \( u(t) \), \( u'(t) \), \( v(t) \), and \( v'(t) \) into the product rule formula:\[ y' = (-2t)\coth^{-1}(t) + (1-t^2)\left(\frac{-1}{1-t^2}\right) \]Simplify the expression:\[ y' = -2t\coth^{-1}(t) - 1 \] (since \( (1-t^2)\frac{-1}{1-t^2} = -1 \)).

Step 5: Simplify the Derivative Expression

The final expression for the derivative is:\[ y' = -2t\coth^{-1}(t) - 1 \]. This is the derivative of \( y \) with respect to \( t \).

Key Concepts

Product Rule in Differentiation

The product rule is a fundamental differentiation technique used when dealing with the derivative of a product of two functions. If you have a function in the form of \( y = u(t) v(t) \), where both \( u(t) \) and \( v(t) \) are differentiable functions, the product rule comes to the rescue. The rule states that the derivative \( y' \) is given by:

\( y' = u'(t) v(t) + u(t) v'(t) \)

To apply this, follow these simple steps:

• First, differentiate \( u(t) \) to get \( u'(t) \).
• Next, differentiate \( v(t) \) to obtain \( v'(t) \).
• Lastly, substitute all these components into the product rule formula.

Using these steps allows for an organized approach to finding the derivative of complex functions that are products of simpler ones.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are similar to inverse trigonometric functions but involve hyperbolic functions like \( \sinh \) and \( \cosh \). One such function is the inverse hyperbolic cotangent, denoted as \( \coth^{-1}(x) \). This function represents the inverse of the hyperbolic cotangent function.

The derivative of \( \coth^{-1}(t) \) is given by:

• \( \frac{d}{dt} \coth^{-1}(t) = \frac{-1}{1-t^2} \)

It's important to remember this derivative because it often appears in problems where differentiation involves inverse hyperbolic functions.

These derivative formulas are derived from the relationships that define inverse hyperbolic functions and are crucial for solving calculus problems effectively.

Differentiation Techniques

Differentiation techniques encompass a variety of methods needed to find the derivative of functions. Beyond basic rules like the product rule or chain rule, many problems require knowledge of derivatives of specific functions, including trigonometric, exponential, and hyperbolic functions.

To effectively differentiate functions:

• Identify the form of the function: product, quotient, chain, or a mix.
• Choose the appropriate rule or combination of rules.
• Recognize special functions like inverse trigonometric or hyperbolic functions.
• Write out necessary derivatives separately before substitution.

Developing these techniques hones your ability to tackle a range of calculus problems, allowing you to differentiate complex functions using a structured approach.