Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=\sqrt[4]{2 x+1}\)

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{2}(2x + 1)^{-3/4}\).

Step 1: Understand the Expression

We are given the function \(y = \sqrt[4]{2x + 1}\). This is the same as \(y = (2x + 1)^{1/4}\). Our task is to find the derivative \(\frac{dy}{dx}\).

Step 2: Apply the Power Rule

To differentiate \(y = (2x + 1)^{1/4}\), use the power rule for derivatives, \(\frac{d}{dx}[u^n] = n u^{n-1} \frac{du}{dx}\), where \(u = 2x + 1\) and \(n = \frac{1}{4}\).

Step 3: Differentiate the Inner Function

First, find \(\frac{du}{dx}\) where \(u = 2x + 1\). The derivative of \(2x + 1\) with respect to \(x\) is simply 2.

Step 4: Combine the Derivatives

Using the chain rule, we have \(\frac{dy}{dx} = \frac{1}{4}(2x + 1)^{-3/4} \cdot 2\). Simplify this to get \(\frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-3/4}\).

Step 5: Interpret the Result

The derivative \(\frac{dy}{dx}\) gives the rate of change of \(y\) with respect to \(x\). For our given equation, it is \(\frac{1}{2}(2x + 1)^{-3/4}\).

Key Concepts

Understanding the Power Rule

The power rule is a foundational tool in calculus for finding derivatives of functions raised to a power. It's especially handy when dealing with polynomials or expressions involving a variable raised to any exponent.
The mathematical formula for the power rule is: If you have an expression of the form \(f(x) = x^n\), then its derivative \(f'(x) = nx^{n-1}\).
In this rule, \(n\) is usually a constant. When we differentiate, we bring the exponent \(n\) down in front and subtract 1 from the original power.

• For example, for \(g(x) = x^3\) using the power rule gives you \(g'(x) = 3x^2\).
• Even for negative or fractional powers, the power rule holds true. For instance, \(h(x) = x^{-1}\) has a derivative of \(h'(x) = -x^{-2}\).

Remembering the power rule helps you quickly differentiate many functions without having to repeatedly apply basic derivative definitions from scratch.

Why the Chain Rule is Essential

The chain rule is essential when dealing with composite functions, where one function is nested inside another. This is common when you're faced with expressions such as \((u(x))^n\). The chain rule provides a strategy for differentiating these more complex expressions.
The chain rule states that the derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). In simpler terms:

• Identify the outer and inner functions. If \(y = (2x + 1)^{1/4}\), here \(u = 2x + 1\) is the inner function and the outer function is \(u^{1/4}\).
• Differentiate the outer function leaving the inner function as it is, and then multiply by the derivative of the inner function.

Using the chain rule allows us to handle the situation when the argument of the power function, or the inner function, is more than just a simple variable.

Step-by-Step Derivative Calculation

Calculating derivatives can often seem like a multi-step process, especially if the function is not simple. The key is to break it down by understanding and applying the rules.First, rewrite the function if necessary to a more workable form. In our case, \(y = \sqrt[4]{2x + 1}\) was rewritten as \(y = (2x + 1)^{1/4}\). This transformation simplifies the use of the power rule.
Next, identify the need for the chain rule. Recognize that you have a function of a function — an "outer function" raised to a power, and an "inner function" (\(2x + 1\)). Then, apply the power rule to differentiate the outer function: bring down the exponent (\(1/4\)) and reduce its power by 1 (\(-3/4\)). Combine this with the derivative of the inner function, which is calculated separately (in this instance, the derivative of \(2x + 1\) is simply 2).

• Multiply the results to get the total derivative: \(\frac{dy}{dx} = \frac{1}{4}(2x + 1)^{-3/4} \cdot 2\).
• Simplify wherever possible to make the expression clearer, resulting in \(\frac{dy}{dx} = \frac{1}{2}(2x + 1)^{-3/4}\).

This combination of understanding the necessary rules and applying them step by step makes derivative calculation more approachable and intuitive.