Problem 20 - Thomas Calculus

Question

In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\left(x^{2} \ln x\right)^{4} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 4(x^2 \ln x)^3 (2x \ln x + x)\)

Step 1: Recognize the Outer Function

In this function, we need to differentiate the expression \(y = (x^2 \ln x)^4\). Notice that this is a composition of functions, where an outside function \(u^4\) is applied to \(u = x^2 \ln x\). We will use the chain rule to differentiate this.

Step 2: Apply the Chain Rule

The chain rule tells us that \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). First, identify \(u\) and then differentiate \(y\) with respect to \(u\) and \(u\) with respect to \(x\).

Step 3: Differentiate the Outer Function

For \(y = u^4\), we have \(\frac{dy}{du} = 4u^3\). Substitute back \(u = x^2 \ln x\) into the expression, so \(\frac{dy}{du} = 4(x^2 \ln x)^3\).

Step 4: Differentiate the Inner Function

Now, find \(\frac{du}{dx}\) for \(u = x^2 \ln x\). Use the product rule: \(\frac{d}{dx}(x^2 \ln x) = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x)\).

Step 5: Apply the Product Rule

Calculate \(\frac{d}{dx}(x^2) = 2x\) and \(\frac{d}{dx}(\ln x) = \frac{1}{x}\). Using the product rule:\[\frac{du}{dx} = 2x \cdot \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x\].

Step 6: Combine Results

Now substitute the expressions from Steps 3 and 5 back into the chain rule:\[\frac{dy}{dx} = 4(x^2 \ln x)^3 \times (2x \ln x + x)\].

Key Concepts

Chain Rule

In calculus, the chain rule is a formula used to find the derivative of a composite function. When you have a function composed of two or more functions, the chain rule helps you differentiate by breaking down the process into simpler steps. So, instead of getting overwhelmed by the complexity of a function, you can handle each smaller part independently.
For example, consider a composite function, like the one in the exercise, where we have \(y = (x^2 \ln x)^4\). Here, we can think of two functions: the outer function \(u^4\) and the inner function \(u = x^2 \ln x\). The key to using the chain rule is to first differentiate the outer function with respect to the inner function, and then multiply it by the derivative of the inner function with respect to \(x\).
This leads us to the chain rule formula: \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). By using this approach, you transform a complex task into a series of simpler, more manageable steps. Understanding this basic rule allows you to take on a variety of problems involving composite functions easily.

Product Rule

When dealing with derivatives, the product rule is a crucial tool for functions that multiply each other. It tells us how to differentiate expressions where two functions are multiplied together. For the function \(u = x^2 \ln x\), we apply the product rule because it is a product of two separate functions: \(x^2\) and \(\ln x\).
The product rule states: if you have two functions, \(f(x)\) and \(g(x)\), their derivative is given by \(f'(x)g(x) + f(x)g'(x)\).

• Differentiating \(x^2\) gives you \(2x\).
• Differentiating \(\ln x\) gives you \(\frac{1}{x}\).

Therefore, applying the product rule:

• \(\frac{d}{dx}(x^2 \ln x) = (2x) \ln x + x^2 \left(\frac{1}{x}\right)\).

This simplifies to \(2x \ln x + x\). By understanding and practicing this rule, you can handle more complex derivatives involving products of functions.

Composite Functions

Composite functions involve applying one function to the result of another function. The exercise exemplifies such a situation with \(y = (x^2 \ln x)^4\). In this case, you deal with a function composed of \(x^2 \ln x\) raised to the power of 4.
Composite functions require you to be familiar with both the inner and outer functions to differentiate them correctly.

• The inner function is \(u = x^2 \ln x\), which needs to be differentiated using the product rule.
• The outer function is \(u^4\), which is straightforward using the power rule, resulting in \(4u^3\).

Once you have these derivatives, they can be combined using the chain rule to find the derivative of the entire composite function. This method of breaking down and analyzing each part helps simplify otherwise complex problems and is essential for mastering derivatives involving composite functions.