Problem 30 - Essential Calculus Early Transcendentals

Question

Find the derivative. Simplify where possible. $$ y=x \operatorname{coth}\left(1+x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( \operatorname{coth}(1 + x^2) - 2x^2 \operatorname{csch}^2(1 + x^2) \).

Step 1: Identify the Derivative Rules Needed

To find the derivative of the function, we need to apply the product rule because we have a product of two functions: \(x\) and \(\operatorname{coth}(1 + x^2)\). Recall that the product rule states \((uv)' = u'v + uv'\).

Step 2: Differentiate the First Function

Differentiate \(x\) with respect to \(x\). We have:\[\frac{d}{dx}[x] = 1\]

Step 3: Differentiate the Second Function

Differentiate \(\operatorname{coth}(1 + x^2)\) using the chain rule. Let \(u = 1 + x^2\), then we need the derivative of \(\operatorname{coth}(u)\) with respect to \(u\), and the derivative of \(u\) with respect to \(x\):\[\frac{d}{du}[\operatorname{coth}(u)] = -\operatorname{csch}^2(u) \]Now differentiate \(u = 1 + x^2\):\[\frac{d}{dx}[1 + x^2] = 2x\]Thus, the derivative \(\frac{d}{dx}[\operatorname{coth}(1 + x^2)]\) is:\[-\operatorname{csch}^2(1 + x^2) \cdot 2x\]

Step 4: Apply the Product Rule

Substitute the derivatives found into the product rule:\[\frac{d}{dx}[x \operatorname{coth}(1 + x^2)] = 1 \cdot \operatorname{coth}(1 + x^2) + x \cdot (-\operatorname{csch}^2(1 + x^2) \cdot 2x)\]

Step 5: Simplify the Expression

Simplify the expression from the product rule:\[\frac{d}{dx}[x \operatorname{coth}(1 + x^2)] = \operatorname{coth}(1 + x^2) - 2x^2 \operatorname{csch}^2(1 + x^2)\]

Key Concepts

Product Rule in Calculus

The product rule is an essential tool in calculus for finding derivatives when dealing with the product of two functions. It helps us calculate the derivative of a product without expanding it. The rule can be stated as follows: if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product \( uv \) is given by:\[(uv)' = u'v + uv'\]To apply the product rule, you:

• Differentiate the first function while keeping the second function the same.
• Then, differentiate the second function while keeping the first function the same.
• Add these two derivatives together.

This rule is particularly helpful when dealing with products where neither function can be easily simplified. For example, consider a function like \( y = x \operatorname{coth}(1 + x^2) \). Here, the product rule lets us take the derivative by considering \( x \) and \( \operatorname{coth}(1 + x^2) \) separately, then combining the results.

Understanding the Chain Rule

The chain rule is a powerful derivative rule used when dealing with composite functions. It allows us to differentiate a function that is nested inside another function. The chain rule can be stated as:If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).To apply this rule, follow these steps:

• Identify the outer function and differentiate it with respect to its argument.
• Next, differentiate the inner function with respect to its variable.
• Multiply these derivatives together.

In the function \( \operatorname{coth}(1 + x^2) \), we see a perfect scenario for applying the chain rule, where \( u = 1 + x^2 \) is the inner function. First differentiate \( \operatorname{coth}(u) \), which gives \( -\operatorname{csch}^2(u) \). Then differentiate the inner function \( u \) resulting in \( 2x \). Multiply these two derivatives together to form the derivative of the composite function.

Hyperbolic Functions and Their Derivatives

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas rather than circles. Some common hyperbolic functions include \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \). Their derivatives follow specific patterns similar to those of trigonometric functions:

• \( \frac{d}{dx}[\sinh(x)] = \cosh(x) \)
• \( \frac{d}{dx}[\cosh(x)] = \sinh(x) \)
• \( \frac{d}{dx}[\tanh(x)] = \operatorname{sech}^2(x) \)
• \( \frac{d}{dx}[\operatorname{coth}(x)] = -\operatorname{csch}^2(x) \)

These functions are often involved in calculus problems where derivatives are needed for hyperbolic expressions. In our example, \( \operatorname{coth}(1 + x^2) \) involves the hyperbolic cotangent, whose derivative is \( -\operatorname{csch}^2(u) \) where \( u \) is the argument. Understanding these derivatives helps solve more complex expressions like the one given in the exercise.