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

\(D_{x} y = \frac{2}{1-(2x-3)^2}\).

Step 1: Understand the Problem

We are asked to find \(D_{x} y\), which means we need to find the derivative of \(y\) with respect to \(x\). Here, \(y = \tanh^{-1}(2x - 3)\). The function \(\tanh^{-1}(u)\) is the inverse hyperbolic tangent function.

Step 2: Recognize the Chain Rule

The expression \(y = \tanh^{-1}(2x - 3)\) is a composition of functions. We have an outer function, \(u = \tanh^{-1}(v)\), and an inner function \(v = 2x - 3\). To find \(D_{x} y\), we will apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx} \).

Step 3: Find the Derivative of the Outer Function

The derivative of \(u = \tanh^{-1}(v)\) with respect to \(v\) is \( \frac{du}{dv} = \frac{1}{1-v^2} \). Applying this to our equation, \( \frac{dy}{dv} = \frac{1}{1-(2x-3)^2} \).

Step 4: Find the Derivative of the Inner Function

The inner function is \(v = 2x - 3\). Its derivative with respect to \(x\) is \( \frac{dv}{dx} = 2 \).

Step 5: Apply the Chain Rule

Applying the chain rule, we have:\[ \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx} = \frac{1}{1-(2x-3)^2} \cdot 2 \]This simplifies to:\[ \frac{dy}{dx} = \frac{2}{1-(2x-3)^2} \]

Step 6: Final Expression

The derivative of \(y = \tanh^{-1}(2x - 3)\) with respect to \(x\) is:\[ \frac{dy}{dx} = \frac{2}{1-(2x-3)^2} \].

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus, particularly useful when dealing with composite functions. Composite functions are functions nested within one another, like Russian dolls.
For example, consider a function defined as \( y = f(g(x)) \), which means \( y \) depends on \( g(x) \). Here, \( f \) is the outer function, and \( g \) is the inner function.

When differentiating such functions, the Chain Rule tells us to multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function. Mathematically, this is expressed as:
\[ \frac{dy}{dx} = \frac{dy}{dg} \times \frac{dg}{dx} \]

• \( \frac{dy}{dg} \) is the derivative of the outer function with respect to the inner function.
• \( \frac{dg}{dx} \) is the derivative of the inner function with respect to \( x \).

This rule helps us find the rate of change of the entire function in relation to \( x \), even when \( y \) isn't directly a function of \( x \). In our exercise, this concept allowed us to effectively differentiate \( \tanh^{-1}(2x - 3) \) by seeing it as a composition of \( \tanh^{-1} \) with \( 2x - 3 \).

Inverse Hyperbolic Functions

Inverse hyperbolic functions like \( \tanh^{-1}(x) \) offer a way to "undo" hyperbolic functions, similar to how inverse logarithmic functions, like \( \ln(x) \), relate to exponential functions.
The hyperbolic tangent function, \( \tanh(x) \), is utilized in various applications, such as curve fitting in data sets and models related to exponential growth or decay.

Holding key properties of inverse hyperbolic functions allows us to rethink strategies when finding derivatives or integrals. The formula for the inverse hyperbolic tangent function \( \tanh^{-1}(x) \) is:

\[ \tanh^{-1}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \text{ for } -1 < x < 1 \]

• Thus, the derivative of \( \tanh^{-1}(v) \) with respect to \( v \) is \( \frac{1}{1-v^2} \).

In our example, understanding this property helped us determine the portion of the derivative related to the composite function's outer layer.

Derivative

Derivatives are a cornerstone of calculus, representing the rate at which a quantity changes. They are incredibly versatile and used in multiple branches of science and engineering to study changes over time or space.
The derivative of a function at a point gives the slope of the tangent line at that point, providing insights into the function's behavior - whether it is increasing, decreasing, or has attained a stationary point like a maximum or minimum.

To differentiate a function, rules like the Power Rule, Product Rule, and Quotient Rule come into play. Importantly, to tackle complex derivatives, the Chain Rule is essential, especially for functions built from nested operations, as seen in composite functions.

• The derivative of a basic variable function \( f(x) = x^n \) is \( f'(x) = nx^{n-1} \).
• The aim is to find the slope of function \( y = \tanh^{-1}(2x - 3) \) by calculating how fast it changes along \( x \).

In our worked example, combining the respective derivatives using the Chain Rule provided us the final derivative expression \( \frac{2}{1-(2x-3)^2} \). This helped us analyze how \( y \) changes with respect to \( x \) through the lens of calculus.