Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\). \(y=\sqrt[4]{3 x^{2}-4 x}\)

Step-by-Step Solution

Answer

\(\frac{d y}{d x} = \frac{6x - 4}{4(3x^2 - 4x)^{3/4}}\)

Step 1: Rewrite the Function

First, let's rewrite the function for easier differentiation. The given function is \(y = \sqrt[4]{3x^2 - 4x}\). We can express it using exponents as \(y = (3x^2 - 4x)^{1/4}\).

Step 2: Apply the Chain Rule

We're dealing with a composite function, so we'll use the chain rule. The chain rule states that if \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Here, \(f(u) = u^{1/4}\) and \(g(x) = 3x^2 - 4x\).

Step 3: Differentiate the Outer Function

Differentiate the outer function \(f(u) = u^{1/4}\) with respect to \(u\). The derivative \(f'(u) = \frac{1}{4}u^{-3/4}\). Thus, the derivative in terms of \(g(x)\) is \(\frac{1}{4}(3x^2 - 4x)^{-3/4}\).

Step 4: Differentiate the Inner Function

Next, differentiate the inner function \(g(x) = 3x^2 - 4x\). The derivative \(g'(x) = 6x - 4\).

Step 5: Combine Using the Chain Rule

Combine the results of the previous steps using the chain rule. Substitute the derivatives into the chain rule formula: \[ \frac{dy}{dx} = \frac{1}{4}(3x^2 - 4x)^{-3/4} \cdot (6x - 4) \].

Step 6: Simplify the Expression

Finally, simplify the expression. \[ \frac{dy}{dx} = \frac{1}{4} (6x - 4)(3x^2 - 4x)^{-3/4} = \frac{6x - 4}{4(3x^2 - 4x)^{3/4}} \].

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus used for computing the derivative of a composite function. When you have a function nested inside another, you need a way to differentiate it. The chain rule provides exactly that.
To understand it, imagine having a function expressed as two layers: an external function and an internal function. For example, in our exercise, we have:

• Outer function: \( f(u) = u^{1/4} \)
• Inner function: \( g(x) = 3x^2 - 4x \)

The chain rule says: If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This means you take the derivative of the outer function, leaving the inner function untouched, then multiply by the derivative of the inner function itself.

Composite Function

A composite function is a function made by combining two functions. It's like stacking two processes, where you apply one function first and then another function to the result of the first.
Our situation here is a perfect example. We start with a simple quadratic expression, \( g(x) = 3x^2 - 4x \), and then raise that expression to the power of \( 1/4 \), represented as \( f(u) = u^{1/4} \).
With composite functions, you are essentially "composing" one function inside another. Understanding how these functions combine and how to differentiate them using the chain rule is crucial in learning calculus. By breaking them into their parts, it becomes easier to see how to tackle their derivatives.

Derivatives

Derivatives are a fundamental concept in calculus. They provide the rate at which a function changes. In simple terms, derivatives tell us the slope of the function at any given point.
For instance, if we look at the derivative of the outer function in our example, \( f(u) = u^{1/4} \), its derivative \( f'(u) \) is \( \frac{1}{4}u^{-3/4} \).
Similarly, for the inner function, \( g(x) = 3x^2 - 4x \), its derivative \( g'(x) \) is \( 6x - 4 \).
Derivatives help us understand the behavior of functions, whether they are increasing, decreasing, or remaining constant at various points.

Simplifying Expressions

After finding the derivative, simplifying the resulting expression makes it more understandable and cleaner to work with.
In our problem, once we applied the chain rule, we ended up with an expression like \( \frac{1}{4} (6x - 4)(3x^2 - 4x)^{-3/4} \). Simplifying this to \( \frac{6x - 4}{4(3x^2 - 4x)^{3/4}} \) makes it more elegant and easier to analyze.
Simplified expressions allow us to see the relationship between the variables more clearly, making predictions and further calculations much more straightforward. It's a vital part of solving calculus problems efficiently.