Problem 47 - Calculus Early Transcendentals

Question

Evaluate the following derivatives. \(f(x)=\cosh ^{-1} 4 x\)

Step-by-Step Solution

Answer

Answer: The derivative of the function \(f(x) = \cosh^{-1}(4x)\) is \(\frac{d}{dx}f(x) = \frac{4}{\sqrt{16x^{2} - 1}}\).

Step 1: Find the derivative of the outer function with respect to \(u\)

The outer function is \(\cosh^{-1}(u)\). The derivative of the inverse hyperbolic cosine function, denoted as \(\frac{d}{du}\cosh^{-1}(u)\), is given by the formula: $$\frac{d}{du}\cosh^{-1}(u) = \frac{1}{\sqrt{u^{2} - 1}}$$

Step 2: Find the derivative of the inner function with respect to \(x\)

The inner function is \(4x\). The derivative of the function with respect to \(x\), denoted as \(\frac{d}{dx}(4x)\), is straightforward: $$\frac{d}{dx}(4x) = 4$$

Step 3: Apply the chain rule

To find the derivative of the entire function, \(f(x)=\cosh^{-1}(4x)\), we will apply the chain rule by multiplying the derivatives of the outer and inner functions: $$\frac{d}{dx}f(x) = \frac{d}{du}\cosh^{-1}(u) \cdot \frac{d}{dx}(4x)$$ Plugging in the results from Steps 1 and 2: $$\frac{d}{dx}f(x) = \frac{1}{\sqrt{u^{2} - 1}} \cdot 4$$ Now, we need to substitute the inner function back in place of \(u\): $$\frac{d}{dx}f(x) = \frac{1}{\sqrt{(4x)^{2} - 1}} \cdot 4$$

Step 4: Simplify the expression

Let's simplify the expression by squaring \((4x)^{2}\): $$\frac{d}{dx}f(x) = \frac{1}{\sqrt{16x^{2} - 1}} \cdot 4$$ Now, let's multiply by 4: $$\frac{d}{dx}f(x) = \frac{4}{\sqrt{16x^{2} - 1}}$$ So, the derivative of the function \(f(x) = \cosh^{-1}(4x)\) is: $$\boxed{\frac{d}{dx}f(x) = \frac{4}{\sqrt{16x^{2} - 1}}}$$

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, especially when dealing with composite functions. It allows us to differentiate a function that is made up of other functions. To apply the chain rule, identify two main components: the outer function and the inner function. The rule states that to find the derivative of the composite function, you must:

• Differentiate the outer function with respect to the inner function.
• Differentiate the inner function with respect to its variable.
• Finally, multiply these two derivatives together.

In our example, the function is given as \(f(x) = \cosh^{-1}(4x)\). Here, \(\cosh^{-1}(u)\) acts as the outer function and \(4x\) is the inner function. By using the chain rule, we take the derivative of \(\cosh^{-1}(u)\) first and then multiply it by the derivative of \(4x\). This careful multiplication gives us the derivative of the entire composite function.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as \(\cosh^{-1}(u)\), are along the lines of their trigonometric counterparts. Much like \(\arccos(u)\), \(\cosh^{-1}(u)\) (read as the inverse hyperbolic cosine of \(u\)) serves to reverse the hyperbolic cosine function, \(\cosh(u)\).For these functions, specific derivative formulas exist. The derivative of the inverse hyperbolic cosine, \(\cosh^{-1}(u)\), is \(\frac{1}{\sqrt{u^2 - 1}}\). This formula emerges because of the underlying properties of hyperbolic functions, which involve exponential growth and decay.
Inverse hyperbolic functions are invaluable in mathematics, physics, and engineering. They can model real-world phenomena, such as electrical circuits and fluid dynamics, where wave-like behavior is present. This highlights the importance of having a firm grasp of their derivatives and behaviors.

Derivative Calculations

Calculating derivatives is central to calculus. It's all about finding the rate at which a function changes. For our function \(f(x) = \cosh^{-1}(4x)\), we strategically break down the process:1. **Differentiate the outer function:** Given \(\cosh^{-1}(u)\), its derivative is \(\frac{1}{\sqrt{u^2-1}}\).2. **Differentiate the inner function:** For \(4x\), the derivative is simply \(4\).3. **Apply the chain rule:** Multiply these derivatives to find the derivative of the composite function.4. **Substitute back and simplify:** Replace \(u\) with \(4x\) in the expression, hence: \([\sqrt{(4x)^2 - 1}]\), simplifies the denominator. Combine it with the constant multiplier \(4\).After these steps, you arrive at the final derivative: \(\frac{4}{\sqrt{16x^2 - 1}}\). Each step matters, underscoring the intricate yet systematic nature of derivative calculations. Understanding the theory behind these steps not only aids in solving problems but also builds a foundational skill set for more advanced mathematical concepts.