Problem 24 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\left(4 x^{2}-1\right) \operatorname{csch}(\ln 2 x) $$

Step-by-Step Solution

Answer

The derivative is \( y' = 8x \operatorname{csch}(\ln(2x)) - \frac{4x^2 - 1}{x} \cdot \operatorname{csch}(\ln(2x))\operatorname{coth}(\ln(2x)) \).

Step 1: Identify the Rule for Differentiation

The given function is a product of two functions: \[ y = f(x) \cdot g(x) \]where \( f(x) = 4x^2 - 1 \) and \( g(x) = \operatorname{csch}(\ln(2x)) \). Therefore, we need to use the product rule for differentiation, which is:\[ (u \cdot v)' = u'v + uv' \] where \( u = f(x) \) and \( v = g(x) \).

Step 2: Differentiate the First Function

Differentiate \( f(x) = 4x^2 - 1 \) with respect to \( x \):\[ f'(x) = \frac{d}{dx}(4x^2 - 1) = 8x \]

Step 3: Differentiate the Second Function

Now, differentiate \( g(x) = \operatorname{csch}(\ln(2x)) \). Start by using the chain rule:First find the derivative of \( \operatorname{csch}(u) \), where \( u = \ln(2x) \):\[ \frac{d}{du}(\operatorname{csch}(u)) = -\operatorname{csch}(u)\operatorname{coth}(u) \]Then differentiate \( u = \ln(2x) \) with respect to \( x \):\[ \frac{d}{dx}(\ln(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \]Combine these results with the chain rule:\[ \frac{d}{dx}(\operatorname{csch}(\ln(2x))) = -\operatorname{csch}(\ln(2x))\operatorname{coth}(\ln(2x)) \cdot \frac{1}{x} \]

Step 4: Apply the Product Rule

Use the derivatives found in Steps 2 and 3 in the product rule:\[ y' = f'(x) \cdot g(x) + f(x) \cdot g'(x) \]Substitute the values:\[ y' = (8x) \cdot \operatorname{csch}(\ln(2x)) + (4x^2 - 1) \cdot \left(-\operatorname{csch}(\ln(2x))\operatorname{coth}(\ln(2x)) \cdot \frac{1}{x}\right) \]

Step 5: Simplify the Expression

Simplify the expression for the derivative:\[ y' = 8x \operatorname{csch}(\ln(2x)) - (4x^2 - 1) \cdot \operatorname{csch}(\ln(2x))\operatorname{coth}(\ln(2x)) \cdot \frac{1}{x} \]This is the derivative \( y' \).

Key Concepts

Product Rule

When we need to differentiate a product of two functions, the product rule is our go-to method. Suppose we have two functions, say, \( f(x) \) and \( g(x) \). The product rule states that the derivative of the product \( f(x) \cdot g(x) \) is given by:

• \( (f(x) \cdot g(x))' = f'(x) \cdot g(x) + f(x) \cdot g'(x) \)
• Essentially, we take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function.

In the example problem, we identified \( f(x) = 4x^2 - 1 \) and \( g(x) = \operatorname{csch}(\ln(2x)) \). By applying the product rule, we could systematically find the derivative of these complex functions multiplied together.

Chain Rule

The chain rule is a powerful tool for finding the derivative of a composite function. A composite function is where one function is applied inside another function, like \( g(f(x)) \). To differentiate composite functions, we use the chain rule which states:

• \( \frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x) \)
• This means we take the derivative of the outer function while maintaining the inner function as it is, and then multiply by the derivative of the inner function.

For instance, in differentiating \( \operatorname{csch}(\ln(2x)) \), we first treat the \( \ln(2x) \) inside the \( \operatorname{csch} \) function as a separate function \( u \). Next, we find \( \frac{d}{dx}(\ln(2x)) \) and multiply it with the derivative of \( \operatorname{csch}(u) \) using the chain rule. This step is crucial for dealing with nested functions like these.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola instead of a circle. They have similar names, such as hyperbolic sine \( \sinh(x) \), hyperbolic cosine \( \cosh(x) \), and hyperbolic cosecant \( \operatorname{csch}(x) \), among others. These functions have unique properties and identities.

• Key properties include: \( \operatorname{csch}(x) = \frac{1}{\sinh(x)} \)
• Derivatives of hyperbolic functions differ slightly from their trigonometric counterparts; for example, \( \frac{d}{dx}[\operatorname{csch}(x)] = -\operatorname{csch}(x)\operatorname{coth}(x) \).

Understanding hyperbolic functions is crucial, especially when dealing with exponential growth situations, and their derivatives emerge frequently in complex calculus problems. Differentiating hyperbolic functions often involves using both the product and chain rules, as demonstrated in the problem solution example.