Problem 9 - College Algebra and Calculus: An Applied Approach

Question

Find \(d y / d u, d u / d x\), and \(d y / d x\). $$ y=u^{2}, u=4 x+7 $$

Step-by-Step Solution

Answer

The derivatives are \(dy/du = 2u\), \(du/dx = 4\), and \(dy/dx = 8u\).

Step 1: Find \(dy/du\)

To find the derivative \(dy/du\), differentiate the equation \(y = u^2\) with respect to \(u\). So, \(dy/du = 2u\).

Step 2: Find \(du/dx\)

Next, differentiate the equation \(u = 4x + 7\) with respect to \(x\) to find \(du/dx\). This yields \(du/dx = 4\).

Step 3: Find \(dy/dx\)

The final step is to find \(dy/dx\). According to the chain rule, \(dy/dx = dy/du * du/dx\). After substituting the previously found derivatives into this formula, we arrive at: \(dy/dx = 2u * 4 = 8u\).

Key Concepts

Understanding Derivatives

Derivatives are a foundational concept in calculus, expressing how a function changes as its inputs change. At its core, a derivative measures the rate at which one quantity changes with respect to another. For example, if we have a function

• \( y = x^2 \)
• Then, the derivative \( dy/dx \) measures how the function \( y \) changes with respect to changes in \( x \)

When finding derivatives:

• Differentiation is the primary process used
• We apply different rules depending on the function’s form

Simple functions, like polynomials, generally have straightforward derivatives, but we'll need rules like the power rule to assist us. In the exercise, the primary derivatives are found for \( y = u^2 \) and \( u = 4x + 7 \), where each brings us closer to understanding how changes in \( x \) affect \( y \) through the intermediary variable \( u \).
The chain rule specifically connects these relationships, allowing us to bridge the gap between variables.

The Method of Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations not easily solved for one variable in terms of another, or when variables are intertwined. Instead of explicitly solving for a variable, we differentiate each term of the equation directly.
In the context of the exercise, implicit differentiation becomes essential when tying together derivatives like \( dy/du \) and \( du/dx \). This method allows us to intuitively apply the chain rule, providing a path to easily link these separate derivatives.

• Identify independent and dependent variables
• Diligently apply differentiation rules

This further solidifies the step-by-step approach:

• Firstly, expressing derivatives \( dy/du \) and \( du/dx \)
• Then, logically connecting them via the chain rule

This yields \( dy/dx = dy/du \cdot du/dx \). In cases where the equation is more complex, implicit differentiation will involve additional algebraic manipulation.

Exploring Function Composition

Function composition involves nesting one function inside another and is crucial when understanding how changes in one variable affect another through an intermediary. Similarly, a composite function is one where the input of one function becomes the output of another.
For the exercise provided, the composition exists between \( y = u^2 \) and \( u = 4x + 7 \). The output of the function \( u \) becomes the input for \( y \), creating a layered relationship. This architecture influences how we approach finding derivatives.

• The chain rule effectively captures this layered dependency
• We break down complex dependencies into manageable parts

We can decompose this process as:

• First, ascertain how \( y \) changes with \( u \) (via \( dy/du \))
• Next, determine how \( u \) changes with \( x \) (using \( du/dx \))
• Finally, connect these to observe the overarching change of \( y \) relative to \( x \) (achieved through \( dy/dx \))

With these insights, function composition succinctly describes dynamic systems where inputs and outputs interconnect fluidly. Applying these principles offers a clearer path to solving more intricate mathematical problems.