Problem 26 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\cosh ^{-1} 2 \sqrt{x+1}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1}{\sqrt{(4x+3)(x+1)}} \)

Step 1: Understand the Function

The function is given as \(y = \cosh^{-1} (2 \sqrt{x+1})\). This involves the inverse hyperbolic cosine function, where \( \cosh^{-1}(u) \) is the inverse hyperbolic cosine of \( u \). Our goal is to find \( \frac{dy}{dx} \).

Step 2: Differentiate Using the Chain Rule

To differentiate \( y = \cosh^{-1} (2 \sqrt{x+1}) \), apply the chain rule which states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), with \( u = 2 \sqrt{x+1} \).

Step 3: Differentiate \(\cosh^{-1}(u)\) with Respect to \(u\)

The derivative of \( \cosh^{-1}(u) \) is \( \frac{1}{\sqrt{u^2 - 1}} \). Hence, \( \frac{dy}{du} = \frac{1}{\sqrt{(2 \sqrt{x+1})^2 - 1}} = \frac{1}{\sqrt{4(x+1) - 1}} = \frac{1}{\sqrt{4x + 3}} \).

Step 4: Differentiate \(2\sqrt{x+1}\) with Respect to \(x\)

For \( u = 2 \sqrt{x+1} \), \( \frac{du}{dx} = 2 \cdot \frac{d}{dx}(\sqrt{x+1}) \). The derivative of \( \sqrt{x+1} \) is \( \frac{1}{2\sqrt{x+1}} \). Hence, \( \frac{du}{dx} = 2 \cdot \frac{1}{2\sqrt{x+1}} = \frac{1}{\sqrt{x+1}} \).

Step 5: Combine the Derivatives

Substitute \( \frac{dy}{du} \) and \( \frac{du}{dx} \) into the chain rule: \(\frac{dy}{dx} = \frac{1}{\sqrt{4x + 3}} \cdot \frac{1}{\sqrt{x+1}} \).

Step 6: Simplify the Expression

Combine the terms: \( \frac{dy}{dx} = \frac{1}{\sqrt{(4x+3)(x+1)}} \). This is the simplified expression for the derivative.

Key Concepts

Chain Rule

In calculus, the chain rule is a powerful technique for calculating derivatives of composite functions. It allows us to differentiate a function that is composed of other functions, easily handling complex expressions by breaking it down into manageable parts. Let's say you have a function composed like this:

• Suppose function \( y = f(u) \) where \( u = g(x) \).
• The chain rule tells us that the derivative \( \frac{dy}{dx} \) is obtained by multiplying the derivative of the outer function \( f(u) \) (with respect to \( u \)), by the derivative of the inner function \( g(x) \) (with respect to \( x \)), i.e., \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

In this exercise, \( y = \cosh^{-1}(2\sqrt{x+1}) \) is such a composite function, given by \( u = 2\sqrt{x+1} \). We first find each derivative separately and then multiply them, aligning with the chain rule approach.

Derivative Calculation

Calculating derivatives involves finding a function that describes the rate of change of one variable with respect to another. For our problem, we need to find \( \frac{dy}{dx} \) where \( y = \cosh^{-1}(2\sqrt{x+1}) \). Using the chain rule, this calculation involves two main derivatives:

• First, we need to find the derivative of the inverse hyperbolic cosine function. The derivative of \( \cosh^{-1}(u) \) is \( \frac{1}{\sqrt{u^2 - 1}} \). So when \( u = 2\sqrt{x+1} \), this becomes \( \frac{1}{\sqrt{(2\sqrt{x+1})^2 - 1}} = \frac{1}{\sqrt{4x + 3}} \).
• Second, we find \( \frac{du}{dx} \) for \( u = 2 \sqrt{x+1} \). That's derived from \( 2 \) times the derivative of \( \sqrt{x+1} \), which is \( \frac{1}{2\sqrt{x+1}} \). Thus, \( \frac{du}{dx} = 2 \times \frac{1}{2\sqrt{x+1}} = \frac{1}{\sqrt{x+1}} \).

Finally, these derivatives are combined to get \( \frac{dy}{dx} = \frac{1}{\sqrt{4x + 3}} \times \frac{1}{\sqrt{x+1}} = \frac{1}{\sqrt{(4x+3)(x+1)}} \). This results in a simplified derivative expression for the original function.

Hyperbolic Cosine

The hyperbolic cosine function, denoted as \( \cosh(x) \), is similar to the regular cosine function but is used in hyperbolic geometry. The hyperbolic functions relate to the geometry of the hyperbola, analogous to how trigonometric functions relate to the circle. For the inverse hyperbolic cosine, \( \cosh^{-1}(u) \), we're dealing with solving for interest in terms of the original hyperbolic function:

• The inverse hyperbolic cosine, \( \cosh^{-1}(u) \), essentially measures the distance along the real axis required to achieve a given hyperbolic angle.
• These functions are pivotal in calculus when understanding complex or logarithmic expressions due to their unique properties and relationships.

In this exercise, recognizing \( \cosh^{-1} \) and how to handle it when deriving helps in applying the chain rule correctly, ensuring you get the right rate of change for the composed function with respect to \( x \).