Problem 2 - University Calculus: Early Transcendentals

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\). $$y=2 u^{3}, \quad u=8 x-1$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 48(8x-1)^2\)

Step 1: Determine Function Derivatives

First, identify the functions involved and their derivatives. The function given is \(y = f(u) = 2u^3\). Differentiate with respect to \(u\) to get \(f'(u) = \frac{d}{du}(2u^3) = 6u^2\). Next, consider \(u = g(x) = 8x - 1\), and differentiate with respect to \(x\) to find \(g'(x) = \frac{d}{dx}(8x - 1) = 8\).

Step 2: Substitute and Apply the Chain Rule

Use the chain rule to find \(\frac{dy}{dx}\). According to the chain rule formula given, \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Substitute \(f'(g(x)) = 6u^2\) with \(u = 8x - 1\), \(f'(g(x)) = 6(8x-1)^2\). The derivative of \(u\) with respect to \(x\) is \(g'(x) = 8\). Combine these to get \(\frac{dy}{dx} = 6(8x-1)^2 \times 8\).

Step 3: Simplify the Expression

Calculate \(\frac{dy}{dx}\) by simplifying the expression. Start with \(\frac{dy}{dx} = 6(8x-1)^2 \times 8\). First, simplify \(6 \times 8 = 48\), resulting in \(\frac{dy}{dx} = 48(8x-1)^2\). This is the final simplified expression for the derivative of \(y\) with respect to \(x\).

Key Concepts

Understanding Derivatives

In calculus, a derivative represents the rate of change of a function with respect to a variable. Think of it like measuring how a road's slope changes while you're driving. The derivative explains how one quantity changes in response to another. For our exercise, we dealt with two functions: \(y = 2u^3\) linked to \(u = 8x - 1\). Each has a derivative:

• The derivative of \(2u^3\) with respect to \(u\) is \(6u^2\).
• The derivative of \(8x-1\) with respect to \(x\) is \(8\).

By finding derivatives of these separate functions, we can see how changes in \(x\) affect \(y\) through \(u\). Derivatives are foundational in calculus, essentially forming the backbone of how we study change.

Function Composition Basics

Function composition essentially involves creating a new function by combining two or more functions together. In our scenario, we have \(y\) as a function of \(u\), and \(u\) itself as a function of \(x\). When we insert one function inside another, like putting one toy inside another, we get the composed function. Here, \(y\) depends on \(x\) through \(u\), making it a perfect example of function composition. In practice, we see:

• \(y = f(u)\) where our original function is \(2u^3\).
• The inner function \(u = g(x)\) is \(8x-1\).

By understanding function composition, we can better grasp how complex relationships between variables work by examining them step-by-step.

Solving a Calculus Exercise with the Chain Rule

Calculus exercises often combine multiple concepts to solve a problem. Here, we focused on the chain rule, a pivotal technique in calculus allowing us to differentiate compositions of functions. Simply put, the chain rule lets us differentiate a function based on its "layers." It's like peeling an onion, layer by layer.For our exercise, it involved:

• Deriving \(f'(u) = 6u^2\) and using \(g(x)\).
• Finding the derivative of \(u = 8x-1\) which is \(g'(x) = 8\).
• Plugging these into the chain rule formula: \(\frac{dy}{dx} = f'(g(x)) \times g'(x)\).
• Substituting and simplifying to yield \(48(8x-1)^2\).

Understanding calculus exercises like this one shows how mathematical techniques are applied to solve real-world problems efficiently.