Problem 231 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for each function. $$y=\left(2 x^{3}-x^{2}+6 x+1\right)^{3}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 3(2x^3 - x^2 + 6x + 1)^2 (6x^2 - 2x + 6) \).

Step 1: Identify the Structure of the Function

We start by recognizing that the function given is of the form \( y = (u(x))^n \) where \( u(x) = 2x^3 - x^2 + 6x + 1 \) and \( n = 3 \). This tells us that we can use the chain rule for differentiation.

Step 2: Apply the Chain Rule

The chain rule states that if \( y = (u(x))^n \), then the derivative \( \frac{dy}{dx} = n imes (u(x))^{n-1} \times \frac{du}{dx} \). In our case, \( n = 3 \), so our formula becomes \( 3 imes (u(x))^2 \times \frac{du}{dx} \).

Step 3: Differentiate the Inner Function

Find \( \frac{du}{dx} \), where \( u(x) = 2x^3 - x^2 + 6x + 1 \). Differentiating term by term, we get: \( \frac{du}{dx} = 6x^2 - 2x + 6 \).

Step 4: Combine the Results

Substitute \( u(x) = 2x^3 - x^2 + 6x + 1 \), \( n = 3 \), and \( \frac{du}{dx} = 6x^2 - 2x + 6 \) into the chain rule formula: \( \frac{dy}{dx} = 3(2x^3 - x^2 + 6x + 1)^2 (6x^2 - 2x + 6) \).

Step 5: Simplify the Expression

The expression \( \frac{dy}{dx} = 3(2x^3 - x^2 + 6x + 1)^2 (6x^2 - 2x + 6) \) can be used as is since there's no further immediate simplification. Hence, \( \frac{dy}{dx} = 3(2x^3 - x^2 + 6x + 1)^2 (6x^2 - 2x + 6) \) is the final answer.

Key Concepts

Chain Rule

Differentiating complex functions can be intimidating, especially when they involve layers, but the chain rule simplifies this process. When we look at a function structured like a nested formula, such as \( y = (u(x))^n \), the chain rule provides a neat mechanism to find its derivative. Basically, the chain rule allows us to differentiate the outer function and then the inner one, linking them through multiplication. For instance, consider \( y = (2x^3 - x^2 + 6x + 1)^3 \). By recognizing the inner function \( u(x) = 2x^3 - x^2 + 6x + 1 \), the chain rule helps us manage this complexity. It tells us that the derivative \( \frac{dy}{dx} \) can be calculated as follows:

• Differentiate the outer function as if the inner function is a single variable.
• Multiply by the derivative of the inner function \( \frac{du}{dx} \).

In the given expression, the power, 3, comes down as a multiplier, reflecting the outer function's differentiation, and the inner function \( u(x) \) remains raised to the power of \( n-1 \). The combined chain results in \( \frac{dy}{dx} = 3(2x^3 - x^2 + 6x + 1)^2 \times (6x^2 - 2x + 6) \), neatly summarizing the use of the chain rule to differentiate this function.

Differentiation Techniques

Differentiation is a primary tool in calculus for finding how a function changes at any point. With numerous techniques available, recognizing the correct one to apply is crucial. For most polynomial and composite functions, the differentiation process involves several steps:

• Identify whether the function is composed of simpler ones, hinting at the chain rule application.
• Use power rule for individual terms in polynomials, multiplying the power by the coefficient and reducing the power by one.
• For complex functions, apply additional rules such as the product rule or quotient rule when needed.

In our exercise, recognizing the function's layers leads to the use of the chain rule. Differentiating each term of \( u(x) = 2x^3 - x^2 + 6x + 1 \) separately, we find its derivative as \( 6x^2 - 2x + 6 \). This approach demonstrates a blend of techniques tailored to the problem type.

Polynomial Differentiation

Polynomials are expressions made up of variables raised to various powers, each multiplied by a coefficient. Differentiating them efficiently involves a consistent process called the power rule. Consider the polynomial function \( u(x) = 2x^3 - x^2 + 6x + 1 \). Here, each term can be differentiated independently using the following steps:

• Multiply the exponent by the coefficient (e.g., the term \( 2x^3 \) gives \( 3 \times 2 = 6 \)).
• Subtract one from the exponent (e.g., \( 3 - 1 = 2 \) for \( 2x^3 \)).
• Apply this method to each term \( x^1 \), and remember that a constant (like 1) differentiates to zero.

Applying these steps, \( \frac{du}{dx} \) results in \( 6x^2 - 2x + 6 \). This straightforward technique of polynomial differentiation sets the stage for handling more complex derivatives, as seen in composite functions. By mastering polynomial differentiation, you'll find tackling advanced calculus problems much smoother.