Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\tanh ^{-1}(2 x-3) $$

Step-by-Step Solution

Answer

\( D_x y = \frac{2}{-(2x-1)(2x-7)} \)

Step 1: Identify the Derivative Needed

The task is to find the derivative of the function \[ y = \tanh^{-1}(2x - 3) \]with respect to \( x \).

Step 2: Recall the Derivative Formula for Inverse Hyperbolic Tangent

The derivative of \( \tanh^{-1}(u) \) with respect to \( u \) is \[ \frac{1}{1-u^2} \]where \( u \) is the inner function. In our case, \( u = 2x - 3 \).

Step 3: Apply the Chain Rule

We have a composite function, using the chain rule:\[ D_x \tanh^{-1}(2x - 3) = \frac{d}{dx} \left( \tanh^{-1}(u) \right) \cdot \frac{du}{dx} \]First, find \( \frac{du}{dx} \) where \( u = 2x - 3 \). The derivative is:\[ \frac{du}{dx} = 2 \].

Step 4: Compute the Derivative

Now, compute the derivative of \( \tanh^{-1}(2x - 3) \):Apply the derivative of the inverse hyperbolic function and the chain rule:\[ D_x \tanh^{-1}(2x - 3) = \frac{1}{1-(2x-3)^2} \cdot 2 \]Simplify the expression. First, calculate:\[ 1 - (2x - 3)^2 = 1 - (4x^2 - 12x + 9) = 1 - 4x^2 + 12x - 9 = -4x^2 + 12x - 8 \].

Step 5: Final Expression for the Derivative

Thus, the derivative is:\[ D_x y = \frac{2}{-4x^2 + 12x - 8} \]Which simplifies to:\[ D_x y = \frac{2}{-(2x-1)(2x-7)} \].

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivatives of composite functions. A composite function can be understood as a function placed inside another function. When you need the derivative of such a function, the chain rule becomes essential. To utilize the chain rule, follow these basic steps:

• Identify the outer function and the inner function. In this case, the outer function is the inverse hyperbolic tangent and the inner function is the linear expression.
• Differentiate the outer function with respect to the inner function.
• Differentiate the inner function with respect to the variable.
• Multiply these derivatives together to find the derivative of the composite function.

In our example, we have the function \(y = \tanh^{-1}(2x - 3)\). Using the chain rule involves differentiating \(\tanh^{-1}(u)\), then multiplying by \(\frac{du}{dx}\), where \(u = 2x - 3\). This calculates as:\[ D_x \tanh^{-1}(2x-3) = \frac{1}{1-(2x-3)^2} \times 2 \]

Derivative of Inverse Functions

Calculating the derivative of inverse functions requires a specific method because they essentially "reverse" the original functions. For inverse hyperbolic functions like, \(y = \tanh^{-1}(x)\), they have their own derivative formulas. The inverse hyperbolic tangent function is represented mathematically as:\[ y = \tanh^{-1}(u) \]Its derivative with respect to \(u\) is given by the formula:\[ \frac{d}{du} \tanh^{-1}(u) = \frac{1}{1-u^2} \]This formula is applicable when finding the derivative of any inverse hyperbolic tangent of a function. To find the derivative with respect to \(x\) instead, we use the chain rule as well.When applied to our specific function \(y = \tanh^{-1}(2x-3)\), the derivative according to the inverse rule is \[ \frac{1}{1-(2x-3)^2} \].With the chain rule, this expression is then multiplied by the derivative of the inner function \(2x-3\), providing the final derivative.

Hyperbolic Tangent

The hyperbolic tangent function, denoted \(\tanh(x)\), is closely related to both exponential functions and geometric hyperbolas. The function is defined as:\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \]This function is analogous to the tangent function in trigonometry but is adapted to hyperbolic geometry. By definition, \(\tanh(x)\) ranges between -1 and 1 and is often used in modeling real-world phenomena where in-between values are common. The inverse hyperbolic tangent or \(\tanh^{-1}(x)\) is the function that "undoes" the actions of \(\tanh(x)\). Its range is all real numbers, making it extremely versatile for solving equations involving hyperbolic relationships. When you differentiate \(\tanh^{-1}(x)\) with respect to \(x\), you access powerful tools for analysis and problem-solving in calculus, especially useful for problems involving hyperbolic identities and transformations. This leads to the derivative: \[ \frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1-x^2} \], This formula is inverted for functions involving transformations, exemplifying the dynamic interchange between hyperbolic and inverse hyperbolic functions.