Problem 41 - Calculus Early Transcendentals

Question

Use a CAS to find \(d y / d x\) $$ y=\left[x \sin 2 x+\tan ^{4}\left(x^{7}\right)\right]^{5} $$

Step-by-Step Solution

Answer

Use the chain rule and derivative rules to find the complex derivative.

Step 1: Identify the Function Structure

The given function is \[y = \left[ x \sin(2x) + \tan^4(x^7) \right]^5.\]It is a composite function, where an outer function raises everything inside the brackets to the 5th power.

Step 2: Apply the Chain Rule

To find the derivative \( \frac{dy}{dx} \), we'll use the chain rule. Firstly, let's denote the inner function as \( u(x) = x \sin(2x) + \tan^4(x^7) \). Thus, \( y = [u(x)]^5 \). The chain rule gives:\[\frac{dy}{dx} = 5[u(x)]^4 \cdot \frac{du}{dx}.\]

Step 3: Find the Derivative of the Inner Function

We need to find \( \frac{du}{dx} \), where \[ u(x) = x \sin(2x) + \tan^4(x^7). \]Using the sum rule and product rule, \[ \frac{du}{dx} = \sin(2x) + 2x \cos(2x) + 4\tan^3(x^7)\cdot\sec^2(x^7)\cdot7x^6. \]Simplifying, we get:\[ \frac{du}{dx} = \sin(2x) + 2x \cos(2x) + 28x^6 \tan^3(x^7) \sec^2(x^7). \]

Step 4: Substitute Back to Find dy/dx

Substituting \( u(x) \) and \( \frac{du}{dx} \) back into the expression derived from the chain rule:\[\frac{dy}{dx} = 5\left[x \sin(2x) + \tan^4(x^7)\right]^4 \left(\sin(2x) + 2x \cos(2x) + 28x^6 \tan^3(x^7) \sec^2(x^7)\right).\]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. It allows us to differentiate functions that are nested within one another.
For example, in the function given, \[y = ig(x \sin 2x + \tan^4(x^7)\big)^5\], we can see two distinct components: the outer function \[f(u) = u^5\], where \[u = x \sin 2x + \tan^4(x^7)\].
The chain rule states that to find the derivative \(\frac{dy}{dx}\), we first differentiate the outer function with respect to the inner function and then multiply it by the derivative of the inner function.
The formula is:

• \[\frac{dy}{dx} = f'(u) \cdot \frac{du}{dx}\]

Applying this principle to our function involves calculating \(f'(u) = 5u^4\) and then multiplying by \(\frac{du}{dx}\). This is effectively applying the chain rule to break down the complexity into manageable parts. Understanding the chain rule is crucial for handling any composite functions in calculus.

Derivative Calculation

Calculating derivatives is one of the core activities in calculus, aimed at finding the rate of change of a function. For our function,\[y = \big(x \sin 2x + \tan^4(x^7)\big)^5\],we first need to find \(\frac{du}{dx}\) before calculating the full \(\frac{dy}{dx}\).
The expression \(u(x) = x \sin 2x + \tan^4(x^7)\) involves a combination of products and powers, requiring the use of both the product rule and the chain rule:

• For \(x \sin(2x)\), we apply the product rule: \(\frac{d}{dx}(x \sin(2x)) = \sin(2x) + 2x \cos(2x)\).
• For \(\tan^4(x^7)\), we apply the chain rule: \(\frac{d}{dx}(\tan^4(x^7)) = 4 \tan^3(x^7) \cdot \sec^2(x^7) \cdot 7x^6\).

By completing these calculations, we achieve the full derivative of the inner function, \(\frac{du}{dx} = \sin(2x) + 2x \cos(2x) + 28x^6 \tan^3(x^7) \sec^2(x^7)\). This part connects directly to finding the derivative of the entire function by utilizing these results within the broader chain rule context.

Composite Functions

A composite function is created when one function is applied within another, which adds layers of complexity to calculus problems. In the given exercise, the function \[y = \big(x \sin 2x + \tan^4(x^7)\big)^5\]is a perfect example of a composite function. Here, the expression \[x \sin 2x + \tan^4(x^7)\] is embedded inside the outer function \(u^5\).
This means the function can be seen as a composition of two functions: an outer function \(y = f(u)\), and an inner function \(u = g(x)\).
Understanding composite functions requires one to systematically address each layer by applying appropriate differentiation rules, such as the chain rule, as mentioned earlier.

• The derivative of a composite function, like in our case, means differentially nesting the outer and inner functions together.
• Appreciating how functions are combined in a composite format helps unearth structure, making the calculation process efficient.

With practice, analyzing and deconstructing composite functions becomes an insightful part of problem-solving in calculus.